Chapter 18 of RD Sharma's Class 11 Mathematics textbook delves into the Binomial Theorem a fundamental concept in the algebra that describes the expansion of the expressions raised to the power. Exercise 18.2 | Set 2 focuses on applying the Binomial Theorem to the solve various types of problems enhancing students' understanding of the theorem's practical applications and its role in the algebraic manipulations.
Binomial Theorem
The Binomial Theorem provides a formula to the expand expressions of the form (a+b)n where a and b are any numbers and n is a non-negative integer. According to the theorem the expansion can be expressed as:
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
where
\binom{n}{k} = \frac{n!}{k!(n-k)!}
This theorem is essential for the solving algebraic problems involving powers and is widely used in the various mathematical and real-world applications.
Question 14. Find the middle terms in the expansion of:
(i) (3x – x3/6)9
Solution:
We have,
(3x – x3/6)9 where, n = 9 (odd number).
So, the middle terms are ((n + 1)/2) = ((9 + 1)/2) = 10/2 = 5 and
((n + 1)/2 + 1) = ((9 + 1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and 6th.
Now,
T5 = T4+1
= 9C4 (3x)9-4 (x3/6)4
=
\frac{9×8×7×6}{5×4×3×2×1}×27×9×\frac{x^{17}}{36×36} =
\frac{189x^{17}}{8} And, T6 = T5+1
= 9C5 (3x)9-5 (x3/6)5
=
-\frac{9×8×7×6}{5×4×3×2×1}×81×9×\frac{x^{19}}{216×36} =
-\frac{21x^{19}}{16}
(ii) (2x2 – 1/x)7
Solution:
We have,
(2x2 – 1/x)7 where, n = 7 (odd number).
So the middle terms are ((n + 1)/2) = ((7 + 1)/2) = 8/2 = 4 and
((n + 1)/2 + 1) = ((7 + 1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5
The terms are 4th and 5th.
Now,
T4 = T3+1
= 7C3 (2x2)7-3 (-1/x)3
= -\frac{7×6×5}{3×2}×16x^8×\frac{1}{x^3}
= − 560 x5
And, T5 = T4+1
= 7C4 (2x2)7-4 (-1/x)4
=
\frac{7×6×5}{3×2}×8×x^6×\frac{1}{x^4} = 280 x2
(iii) (3x – 2/x2)15
Solution:
We have,
(3x – 2/x2)15 where, n = 15 (odd number)
So the middle terms are ((n + 1)/2) = ((15 + 1)/2) = 16/2 = 8 and
((n + 1)/2 + 1) = ((15 + 1)/2 + 1) = (16/2 + 1) = (8 + 1) = 9
The terms are 8th and 9th.
Now,
T8 = T7+1
= 15C7 (3x)15-7 (– 2/x2)7
=
-\frac{15×14×13×12×11×10×9}{7×6×5×4×3×2×1}×3^8×2^7×x^{8-14} =
\frac{-6435×3^8×2^7}{x^{6}} And, T9 = T8+1
= 15C8 (3x)15-8 (– 2/x2)8
=
-\frac{15×14×13×12×11×10×9}{7×6×5×4×3×2×1}×3^7×2^8×x^{7-16} =
\frac{6435×3^7×2^8}{x^{9}}
(iv) (x4 – 1/x3)11
Solution:
We have, (x4 – 1/x3)11 where, n = 11 (odd number)
So the middle terms are ((n + 1)/2) = ((11 + 1)/2) = 12/2 = 6 and
((n + 1)/2 + 1) = ((11 + 1)/2 + 1) = (12/2 + 1) = (6 + 1) = 7
The terms are 6th and 7th.
Now,
T6 = T5+1
= 11C5 (x4)11-5 (1/x3)5
=
- \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2} \times \left( x \right)^{24 - 15} = -462 x9
And, T7 = T6+1
= 11C6 (x4)11-6 (1/x3)6
=
\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2} \left( x \right)^{20 - 18} = 462 x2
Question 15. Find the middle terms in the expansion of:
(i) (x – 1/x)10
Solution:
We have,
(x – 1/x)10 where, n = 10 (even number)
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6th term
Now,
T6 = T5+1
= 10C5 (x)10-5 (–1/x)5
=
-\frac{10×9×8×7×6}{5×4×3×2×1} = −252
(ii) (1 - 2x + x2)n
Solution:
We have, (1 - 2x + x2)n
= (1 - x)2n
Here, n is an even number.
So the middle terms is 2n/2 + 1 = (n + 1)th term
Now,
Tn+1 = 2nCn (-1)n xn
=
\frac{(2n)!}{(n! )^2}( - 1 )^n x^n
(iii) (1 + 3x + 3 x2 + x3)2n
Solution:
We have, (1 + 3x + 3 x2 + x3)2n
= (1 + x )6n
Here, n is an even number.
So the middle terms is (6n/2 + 1) = (3n + 1)th term
Now,
T3n+1 = 6nC3n x3n
=
\frac{(6n)!}{(3n! )^2} x^{3n}
(iv) (2x – x2/4)9
Solution:
We have,
(2x – x2/4)9 where, n = 9 (odd number)
So the middle terms are ((n + 1)/2) = ((9 + 1)/2) = 10/2 = 5 and
((n + 1)/2 + 1) = ((9 + 1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and 6th.
Now,
T5 = T4+1
= 9C4 (2x)9-4 (–x2/4)4
=
\frac{9×8×7×6}{4×3×2×1}×\frac{2^5}{4^4}×x^{5+8} =
\frac{63}{4}x^{13} And, T6 = T5+1
= 9C5 (2x)9-5 (–x2/4)5
=
-\frac{9×8×7×6}{4×3×2×1}×\frac{2^4}{4^5}×x^{4+10} =
-\frac{63}{32}x^{14}
(v) (x - 1/x)2n+1
Solution:
We have, (x - 1/x)2n+1
Here, 2n + 1 is an odd number.
So the middle terms are((2n + 1 + 1)/2)th and ((2n + 1 + 1)/2 + 1)th terms.
The terms are (n + 1)th and (n + 2)th
Now
Tn+1 =
^{2n + 1}{}{C}_n x^{2n + 1 - n} \times \frac{( - 1 )^n}{x^n} =(-1)n 2n+1Cn x
And Tn+2 = Tn+1+1
=
^{2n + 1}{}{C}_{n + 1} x^{2n + 1 - n - 1} \frac{( - 1 )^{n + 1}}{x^{n + 1}} =
(- 1)^{n + 1} \\^{2n + 1}C_{n + 1} \times \frac{1}{x}
(vi) (x/3 + 9y)10
Solution:
We have,
(x/3 + 9y)10 where, n = 10 (even number)
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6th term.
Now,
T6 = T5+1
= 10C5 (x/3)10-5 (9y)5
=
\frac{109876}{54323^5}9^5x^5y^5 = 61236 x5 y5
(vii) (3 – x3/6)7
Solution:
We have,
(3 – x3/6)7 where, n = 7 (odd number).
So the middle terms are ((n + 1)/2) = ((7 + 1)/2) = 8/2 = 4 and
((n + 1)/2 + 1) = ((7 + 1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5
The terms are 4th and 5th.
Now,
T4 = T3+1
= 7C3 (3)7-3 (-x3/6)3
=
-\frac{765}{321}3^4\frac{x^9}{216} =
-\frac{105}{8}x^9 And, T5 = T4+1
= 9C4 (3)9-4 (-x3/6)4
=
\frac{765}{321}3^5\frac{x^{12}}{6^4} =
\frac{35}{48}x^{12}
(viii) (2ax – b/x2)12
Solution:
We have,
(2ax – b/x2)12 where, n = 12 (even number).
So the middle terms are (n/2 + 1) = (12/2 + 1) = 7th term
Now,
T7 = T6+1
= 12C6 (2ax)12-6 (-b/x2)6
= 12C6 (2ax)6 (b/x2)6
= 12C6 (26a6x6)(b6/x12)
= 12C6 (26a6b6/x6)
(ix) (p/x + x/p)9
Solution:
We have,
(p/x + x/p)9 where, n = 9 (odd number).
So the middle terms are ((n + 1)/2) = ((9 + 1)/2) = 10/2 = 5 and
((n + 1)/2 + 1) = ((9 + 1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and 6th.
Now,
T5 = T4+1
= 9C4 (p/x)9-4 (x3/p)4
= 9C4 (p/x)5 (x/p)4
= 9C4 (p/x)
And, T6 = T5+1
= 9C5 (p/x)9-5 (x/p)5
= 9C5 (p/x)4(x/p)5
= 9C5 (x/p)
(x) (x/a – a/x)10
Solution:
We have,
(x/a – a/x)10 where, n = 10 (even number).
So the middle terms are (n/2 + 1) = (10/2 + 1) 6th term
Now,
T6 = T5+1
= 10C5 (x/a)10-5 (-a/x)5
= -10C5 (x/a)5 (a/x)5
= -10C5
= -252
Question 16. Find the term independent of x in the expansion of the following expressions:
(i) (3/2 x2 – 1/3x)9
Solution:
We have,
(3/2 x2 – 1/3x)9
We know, the (r+1)th term of the expression is given by,
Tr+1 = nCr xn-r ar
= 9Cr (3/2x2)9-r (-1/3x)r
=
(-1)^r (^9C_r) (\frac{3^{9-2r}}{2^{9-r}}) x^{18-2r-r} For this term to be independent of x, we must have
=> 18 – 3r = 0
=> 3r = 18
=> r = 18/3
=> r = 6
So, the required term is 7th term.
So, T7 = T6+1
= 9C6 × (39-12)/(29-6)
=
\frac{9×8×7}{3×2}× 3^{-3} × 2^{-3} = 7/18
Hence, the term independent of x is 7/18.
(ii) (2x + 1/3x2)9
Solution:
We have,
(2x + 1/3x2)9
We know, the (r+1)th term of the expression is given by,
Tr+1 = nCr xn-r ar
= 9Cr (2x)9-r (1/3x2)r
=
^9C_r(\frac{2^{9-r}}{3^r})x^{9-r-2r} For this term to be independent of x, we must have
=> 9 – 3r = 0
=> r = 3
So, the required term is 4th term.
So, T7 = T6+1
= 9C6 × (29-3)/(33)
=
\frac{9×8×7}{3×2×1}×\frac{64}{27} = 5376/27
(iii) (2x2 – 3/x3)25
Solution:
We have,
(2x2 – 3/x3)25
We know, the (r+1)th term of the expression is given by,
Tr+1 = nCr xn-r ar
= 25Cr (2x2)25-r (-3/x3)r
= (-1)r 25Cr × 225-r × 3r x50-2r-3r
For this term to be independent of x, we must have
=> 50 – 5r = 0
=> 5r = 50
=> r = 10
So, the required term is 11th term.
So, T11 = T10+1
= (-1)10 25C10 × 225-10 × 310
= 25C10 (215 × 310)
(iv) (3x – 2/x2)15
Solution:
We have,
(3x – 2/x2)15
We know, the (r+1)th term of the expression is given by,
Tr+1 = nCr xn-r ar
= 15Cr (3x)15-r (-2/x2)r
= (-1)r 15Cr × 315-r × 2r x15-r-2r
For this term to be independent of x, we must have
=> 15 – 3r = 0
=> 3r = 15
=> r = 5
So, the required term is 6th term.
So, T6 = T5+1
= (-1)5 15C5 × 315-5 × 25
= −3003 × 310 × 25
(v) (\sqrt{\frac{x}{3}} + \frac{3}{2 x^2})^{10}
Solution:
We have
(\sqrt{\frac{x}{3}} + \frac{3}{2 x^2})^{10} We know, the (r+1)th term of the expression is given by,
Now,
Tr+1 =
^{10}{}{C}_r \left( \sqrt{\frac{x}{3}} \right)^{10 - r} \left( \frac{3}{2 x^2} \right)^r =
^{10}{}{C}_r . \frac{3^{r - \frac{10 - r}{2}}}{2^r} x^\frac{10 - r}{2} - 2r For this term to be independent of x, we must have
=> (10-r)/2 - 2r = 0
=> 10 - 5r = 0
=> r = 2
So, the required term is 3th term.
So, T3 = T2+1
T3 =
^{10}{}{C}_2 \times \frac{3^{2 - \frac{10 - 2}{2}}}{2^2} =
\frac{10 \times 9}{2 \times 4 \times 9} = 5/4
(vi) \left( x - \frac{1}{x^2} \right)^{3n}
Solution:
We have
\left( x - \frac{1}{x^2} \right)^{3n} We know, the (r+1)th term of the expression is given by,
Now,
Tr+1 =
^{3n}{}{C}_r x^{3n - r} \left( \frac{- 1}{x^2} \right)^r = (-1)r 3nCr x3n-r-2r ar
For this term to be independent of x, we must have}
=> 3n - 3r = 0
=> r = n
So, the required term is (n + 1)th term.
So,
Tn+1 = (-1)n 3nCn
(vii) \left( \frac{1}{2} x^{\frac{1}{3}} + x^{\frac{- 1}{5}} \right)^8
Solution:
We have
\left( \frac{1}{2} x^{\frac{1}{3}} + x^{\frac{- 1}{5}} \right)^8 We know, the (r+1)th term of the expression is given by,
Now,
Tr+1 =
^{8}{}{C}_r \left( \frac{1}{2} x^{\frac{1}{3}} \right)^{8 - r} ( x^{\frac{- 1}{5}} )^r =
^{8}{}{C}_r . \frac{1}{2^{8 - r}} x^\frac{8 - r}{3} - \frac{r}{5} For this term to be independent of x, we must have
=> (8 - r)/3 - r/5 = 0
=> 40 - 5r - 3r = 0
=> 8r = 40
=> r = 5
So, the required term is 6th term.
So, T6 = T5+1
T6 =
^{8}{}{C}_5 \times \frac{1}{2^{8 - 5}} =
\frac{8 \times 7 \times 6}{3 \times 2 \times 8} = 7
(viii) (1 + x + 2x3) (3/2x2 – 3/3x)9
Solution:
We have
(1 + x + 2x3) (3/2x2 – 3/3x)9
We know, the (r+1)th term of the expression is given by,
Now,
Tr+1 = (1 + x + 2x3) [(3/2x2) - 9C1 (3/2x2)8 (1/3x) + . . . - 9C7 (3/2x2)2 (1/3x)7]
For this term to be independent of r, we must have
= 9C6 (33/23) (1/36) - 2x3 9C7 (23/33) (1/37) (1/x3)
= 7/18 – 2/27
= (189 – 36)/486
= 153/486
= 17/54
(ix) \left( \sqrt[3]{x} + \frac{1}{2 \sqrt[3]{x}} \right)^{18} , x > 0
Solution:
We have
\left( \sqrt[3]{x} + \frac{1}{2 \sqrt[3]{x}} \right)^{18} We know, the (r+1)th term of the expression is given by,
Now,
Tr+1 =
^{18}{}{C}_r ( x^{1/3} )^{18 - r} \left( \frac{1}{2 x^{1/3}} \right)^r =
^{18}{}{C}_r \times \frac{1}{2^r} x^\frac{18 - r}{3} - \frac{r}{3} For this term to be independent of r, we must have
=> (18 - r)/3 - r/3 = 0
=> 18 - 2r = 0
=> r = 9
So, the required term is 9th term.
So, T9 = T8+1
T9 = 18C9 (1/29)
(x) \left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^6
Solution:
We have
\left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^6 Suppose the (r + 1)th term in the given expression is independent of x.
Now,
Tr+1 =
^{6}{}{C}_r \left( \frac{3}{2} x^2 \right)^{6 - r} \left( \frac{- 1}{3x} \right)^r =
\left( - 1 \right)^r\\^6C_r \times \frac{3^{6 - r - r}}{2^{6 - r}} x^{12 - 2r - r} For this term to be independent of x, we must have
=> 12 - 3r = 0
=> r = 4
Hence, the required term is the 4th term.
So, T4 = T3+1
=
^{6}{}{C}_4 \times \frac{3^{6 - 4 - 4}}{2^{6 - 4}}
= \frac{6 \times 5}{2 \times 1 \times 4 \times 9} = 5/12
Question 17. If the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)18 are equal, find r.
Solution:
We are given,
(1 + x)18
We know, the coefficient of the rth term in the expansion of (1 + x)n is nCr-1.
So, the coefficients of the (2r + 4)th and (r – 2)th terms in the given expansion are,
18C2r+4-1 and 18Cr-2-1
According to the question, we have,
=> 18C2r+4-1 = 18Cr-2-1
=> 18C2r+3 = 18Cr-3
We know if nCr = nCs , then r = s or r + s = n.
=> 2r + 3 = r – 3 or 2r + 3 + r – 3 = 18
=> 2r – r = –3 – 3 or 3r = 18 – 3 + 3
=> r = –6 or 3r = 18
=> r = –6 or r = 6
Ignoring r = – 6 as r cannot be negative.
Therefore, the value of r is 6.
Question 18. If the coefficients of (2r + 1)th term and (r + 2)th term in the expansion of (1 + x)43 are equal, find r.
Solution:
We are given,
(1 + x)43
We know, the coefficient of the rth term in the expansion of (1 + x)n is nCr-1.
So, the coefficients of the (2r + 1)th and (r + 2)th terms in the given expansion are,
43C2r+1-1 and 43Cr+2-1
According to the question, we have,
=> 43C2r+1-1 = 43Cr+2-1
=> 43C2r = 43Cr+1
We know if nCr = nCs , then r = s or r + s = n.
=> 2r = r + 1 or 2r + r + 1 = 43
=> r = 1 or 3r = 42
=> r = 1 or r = 14
Ignoring r = 1 as it gives the same term on both the sides.
Therefore, the value of r is 14.
Question 19. Prove that the coefficient of (r + 1)th term in the expansion of (1 + x)n+1 is equal to the sum of the coefficients of rth and (r+1)th terms in the expansion of (1 + x)n.
Solution:
We know, the coefficients of (r + 1)th term in (1 + x)n+1 is n+1Cr.
So, sum of the coefficients of the rth and (r + 1)th terms in (1 + x)n is,
(1 + x)n = nCr-1 + nCr
As, nCr+1 + nCr = n+1Cr+1
= n+1Cr
Hence proved.
Question 20. Prove that the term independent of x in the expansion of (x + 1/x)2n is \frac{1×3×5×7...(2n-1)}{n!}×2^n .
Solution:
We have,
(x + 1/x)2n
We know the (r + 1)th term is given by,
Tr+1 = nCr xn-r ar
= 2nCr x2n-r (1/x)r
= 2nCr x2n-2r
For this term to be independent of x, we must have,
=> 2n − 2r = 0
=> 2n = 2r
=> r = n
Therefore, the required term is (n+1)th term .
Tn+1 = 2nCn x2n-n (1/x)n
= 2nCn
=
\frac{2n!}{n!n!} =
\frac{2n(2n-1)(2n-2)...5×4×3×2×1}{n!n!} =
\frac{(1×3×5×...(2n-1))×2^n(1×2×3×4...n)}{n!n!} =
\frac{(1×3×5×...(2n-1))×2^n×n!}{n!n!} =
\frac{1×3×5×7...(2n-1)}{n!}×2^n Hence proved.
Question 21. If the coefficients of 5th, 6th and 7th terms of the expansion (1 + x)n are in A.P., find n.
Solution:
We have, (1 + x)n
We know the coefficient of rth term of a binomial expression is given by nCr-1.
Coefficient of 5th term = nC5-1 = nC4
Coefficient of 6th term = nC6-1 = nC5
Coefficient of 7th term = nC7-1 = nC6
According to the question, we have,
=> 2 nC5 = nC4 + nC6
=>
2[\frac{n!}{(n-5)!5!}] = \frac{n!}{(n-4)!4!} + \frac{n!}{(n-6)!6!} =>
\frac{2}{(n-5)!5!}=\frac{1}{(n-4)!4!} + \frac{1}{(n-6)!6!} =>
\frac{2}{(n-5)(n-6)!×5×4!}=\frac{1}{(n-4)(n-5)(n-6)!×4!} + \frac{1}{(n-6)!×6×5×4!} =>
\frac{2}{(n-5)×5}=\frac{1}{(n-4)(n-5)} + \frac{1}{6×5} =>
\frac{2}{5}=\frac{1}{n-4} + \frac{n-5}{30} =>
\frac{2}{5}=\frac{30+n^2-9n+20}{30(n-4)} => 60(n−4) = 150 + 5n2 − 45n + 100
=> 60n − 240 = 250 + 5n2 − 45n
=> 5n2 − 105n + 490 = 0
=> n2 − 21n + 98 = 0
=> n2 − 7n − 14n + 98 = 0
=> n (n − 7) − 14 (n − 7) = 0
=> (n − 7) (n − 14) = 0
=> n = 7 or n = 14
Therefore, the value of n is 7 or 14.
Question 22. If the coefficients of 2nd, 3rd, and 4th terms of the expansion (1 + x)2n are in A.P., show that 2n2 − 9n + 7 = 0.
Solution:
We have, (1 + x)2n
We know the coefficient of rth term of a binomial expression is given by nCr-1.
Coefficient of 2nd term = 2nC2-1 = 2nC1
Coefficient of 3rd term = 2nC3-1 = 2nC2
Coefficient of 4th term = 2nC4-1 = 2nC3
According to the question, we have,
=> 2 2nC2 = 2nC1 + 2nC3
=> 2 =
\frac{^{2n}C_1}{^{2n}C_2}+\frac{^{2n}C_3}{^{2n}C_2} => 2 =
\frac{2}{2n-2+1}+\frac{2n-3+1}{3} =>
\frac{2}{2n-1}+\frac{2n-2}{3} = 2=>
\frac{6+4n^2-4n-2n+2}{3(2n-1)} = 2=> 4n2 − 6n + 8 = 12n − 6
=> 4n2 − 18n + 14 = 0
=> 2 (2n2 − 9n + 7) = 0
=> 2n2 − 9n + 7 = 0
Hence proved.
Question 23. In the expansion of (1 + x)n, the binomial coefficients of three consecutive terms are respectively 220, 495, and 792, find the value of n.
Solution:
We have, (1 + x)n
Let the three consecutive terms be rth, (r+1)th and (r+2)th.
We know the coefficient of rth term of a binomial expression is given by nCr-1.
Coefficient of rth term = nCr-1 = 220
Coefficient of (r+1)th term = nCr+1-1 = nCr = 495
Coefficient of (r+2)th term = nCr+2-1 = nCr+1 = 792
Now,
\frac{^nC_{r+1}}{^nC_r}=\frac{792}{495} =>
\frac{n-(r+1)+1}{r+1}=\frac{792}{495} =>
\frac{n-r}{r+1}=\frac{8}{5} => 5n − 5r = 8r + 8
=> 5n − 13r = 8 . . . . (1)
Also,
\frac{^nC_{r}}{^nC_{r-1}}=\frac{495}{220} =>
\frac{n-r+1}{r}=\frac{9}{4} => 4n − 4r + 4 = 9r
=> 4n − 13r = −4 . . . . (2)
Subtracting (2) from (1), we get,
=> n = 8 + 4
=> n = 12
Therefore, the value of n is 12.
Question 24. If the coefficients of 2nd, 3rd and 4th terms of the expansion (1 + x)n are in A.P., then find the value of n.
Solution:
We have, (1 + x)n
We know the coefficient of rth term of a binomial expression is given by nCr-1.
Coefficient of 2nd term = nC2-1 = nC1
Coefficient of 3rd term = nC3-1 = nC2
Coefficient of 4th term = nC4-1 = nC3
According to the question, we have,
=> 2 nC2 = nC1 + nC3
=>
2 = \frac{^{n}C_1}{^{n}C_2}+\frac{^{n}C_3}{^{n}C_2} =>
2 = \frac{2}{n-2+1}+\frac{n-3+1}{3} =>
\frac{2}{n-1}+\frac{n-2}{3} = 2 =>
\frac{6+n^2-3n+2}{3(n-1)} = 2=> n2 − 3n + 8 = 6 (n − 1)
=> n2 − 3n + 8 = 6n − 6
=> n2 − 9n + 14 = 0
=> n2 − 7n − 2n + 14 = 0
=> n (n−7) − 2 (n−7) = 0
=> (n − 2) (n − 7) = 0
=> n = 2 or n = 7
Ignoring n = 2 as it does not satisfy our condition.
Therefore, the value of n is 7.
Question 25. If in the expansion of (1 + x)n, the coefficients of pth and qth terms are equal, then prove that p + q = n + 2, where p ≠ q.
Solution:
We have, (1 + x)n
We know the coefficient of rth term of a binomial expression is given by nCr-1.
Coefficient of pth term = nCp-1
Coefficient of qth term = nCq-1
According to the question, we have,
=> nCp-1 = nCq-1
=> p − 1 = q − 1 or p − 1 + q − 1 = n
=> p = q or p + q − 2 = n
=> p + q − 2 = n
=> p + q = n + 2
Hence proved.
Question 26. If in the expansion of (1 + x)n, the binomial coefficients of three consecutive terms are respectively 56, 70, and 56, find n and the position of the terms of these coefficients.
Solution:
We have, (1 + x)n
Let the three consecutive terms be rth, (r+1)th and (r+2)th.
We know the coefficient of rth term of a binomial expression is given by nCr-1.
Coefficient of rth term = nCr-1 = 56
Coefficient of (r+1)th term = nCr+1-1 = nCr = 70
Coefficient of (r+2)th term = nCr+2-1 = nCr+1 = 56
Now,
\frac{^nC_{r+1}}{^nC_r}=\frac{56}{70} =>
\frac{n-(r+1)+1}{r+1}=\frac{56}{70} =>
\frac{n-r}{r+1}=\frac{4}{5} => 5n − 5r = 4r + 4
=> 5n − 9r = 4 . . . . (1)
Also,
\frac{^nC_{r}}{^nC_{r-1}}=\frac{70}{56} =>
\frac{n-r+1}{r}=\frac{5}{4} => 4n − 4r + 4 = 5r
=> 4n − r = −4 . . . . (2)
Subtracting (2) from (1), we get,
=> n = 4 + 4
=> n = 8
Putting n = 8 in (1), we get,
=> 5(8) − 9r = 4
=> 40 − 9r = 4
=> 9r = 36
=> r = 4
Therefore, three consecutive terms are 4th, 5th and 6th terms.
Conclusion
Exercise 18.2 | Set 2 of Chapter 18 offers practical problems that reinforce the concepts of the Binomial Theorem. By solving these problems students gain proficiency in the expanding binomials and understanding the theorem's application in the different contexts. Mastery of these exercises enhances algebraic skills and prepares students for the more advanced mathematical concepts.
Are there any common mistakes to avoid when using the Binomial Theorem?
The Common mistakes include the miscalculating the binomial coefficients incorrect application of the formula and overlooking the terms in the expansion. Always double-check calculations and ensure the all terms are included.