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Binomial Theorem question

2023 · 11 Apr · Shift 2 · Q24
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  5. /2023 · 11 Apr · Shift 2 · Q24

Binomial Theorem question

2023 · 11 Apr · Shift 2 · Q24

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
The sum of the coefficients of three consecutive terms in the binomial expansion of (1+x)n+2(1+\mathrm{x})^{\mathrm{n}+2}(1+x)n+2, which are in the ratio 1:3:51: 3: 51:3:5, is equal to :
  1. A
    63
  2. B
    92
  3. C
    25
  4. D
    41
View written solutionFree

Correct answer: A

  1. In the expansion of
(1+x)n+2,(1+x)^{n+2},(1+x)n+2,

the general coefficient is

(n+2r).\binom{n+2}{r}.(rn+2​).
  1. Let the coefficients of three consecutive terms be
(n+2r),(n+2r+1),(n+2r+2).\binom{n+2}{r},\quad \binom{n+2}{r+1},\quad \binom{n+2}{r+2}.(rn+2​),(r+1n+2​),(r+2n+2​).

Given that they are in the ratio

1:3:5.1:3:5.1:3:5.

So we can write

(n+2r):(n+2r+1):(n+2r+2)=1:3:5.\binom{n+2}{r} : \binom{n+2}{r+1} : \binom{n+2}{r+2} = 1:3:5.(rn+2​):(r+1n+2​):(r+2n+2​)=1:3:5.
  1. Use ratios of consecutive binomial coefficients:
(n+2r+1)(n+2r)=n+2−rr+1.\frac{\binom{n+2}{r+1}}{\binom{n+2}{r}} = \frac{n+2-r}{r+1}.(rn+2​)(r+1n+2​)​=r+1n+2−r​.

Since the ratio is 1:31:31:3, we get

n+2−rr+1=3.\frac{n+2-r}{r+1} = 3.r+1n+2−r​=3.

Hence,

n+2−r=3r+3n+2-r = 3r+3n+2−r=3r+3 n−1=4r⇒n=4r+1.n-1 = 4r \quad \Rightarrow \quad n = 4r+1.n−1=4r⇒n=4r+1.
  1. Also,
(n+2r+2)(n+2r+1)=n+1−rr+2.\frac{\binom{n+2}{r+2}}{\binom{n+2}{r+1}} = \frac{n+1-r}{r+2}.(r+1n+2​)(r+2n+2​)​=r+2n+1−r​.

Since the ratio is 3:53:53:5, we get

n+1−rr+2=53.\frac{n+1-r}{r+2} = \frac{5}{3}.r+2n+1−r​=35​.

So,

3(n+1−r)=5(r+2)3(n+1-r) = 5(r+2)3(n+1−r)=5(r+2) 3n+3−3r=5r+103n+3-3r = 5r+103n+3−3r=5r+10 3n−8r=7.3n-8r = 7.3n−8r=7.

Using n=4r+1n=4r+1n=4r+1,

3(4r+1)−8r=73(4r+1)-8r = 73(4r+1)−8r=7 12r+3−8r=712r+3-8r = 712r+3−8r=7 4r=4⇒r=1.4r = 4 \Rightarrow r=1.4r=4⇒r=1.

Thus,

n=4(1)+1=5.n = 4(1)+1 = 5.n=4(1)+1=5.
  1. Therefore the expansion is of
(1+x)n+2=(1+x)7.(1+x)^{n+2} = (1+x)^7.(1+x)n+2=(1+x)7.

The three consecutive coefficients are

(71), (72), (73)=7,21,35,\binom{7}{1},\ \binom{7}{2},\ \binom{7}{3} = 7,21,35,(17​), (27​), (37​)=7,21,35,

which indeed are in the ratio

1:3:5.1:3:5.1:3:5.
  1. Their sum is
7+21+35=63.7+21+35=63.7+21+35=63.

Therefore, the required sum is

63.\boxed{63}.63​.
  1. Comparing with the stored correct answer:
  • Derived answer: 636363
  • Stored correct answer: A (636363)

They match.

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