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Binomial Theorem question

2024 · 31 Jan · Shift 2 · Q58
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Binomial Theorem question

2024 · 31 Jan · Shift 2 · Q58

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
Let the coefficient of xrx^rxr in the expansion of (x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+……….+(x+2)n−1(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots \ldots \ldots .+(x+2)^{n-1}(x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+……….+(x+2)n−1 be αr\alpha_rαr​. If ∑r=0nαr=βn−γn,β,γ∈N\sum_{r=0}^n \alpha_r=\beta^n-\gamma^n, \beta, \gamma \in \mathbb{N}r=0∑n​αr​=βn−γn,β,γ∈N, then the value of β2+γ2\beta^2+\gamma^2β2+γ2 equals ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 25

  1. Write the given expression as a series

The expression is

(x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+⋯+(x+2)n−1.(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\cdots+(x+2)^{n-1}.(x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+⋯+(x+2)n−1.

This is

S=∑k=0n−1(x+3)n−1−k(x+2)k.S=\sum_{k=0}^{n-1}(x+3)^{n-1-k}(x+2)^k.S=k=0∑n−1​(x+3)n−1−k(x+2)k.

Let the coefficient of xrx^rxr in SSS be αr\alpha_rαr​.

  1. Use the sum of coefficients idea

We are given

∑r=0nαr.\sum_{r=0}^{n} \alpha_r.r=0∑n​αr​.

But the sum of coefficients of a polynomial is obtained by putting x=1x=1x=1. So,

∑r=0nαr=S(1).\sum_{r=0}^{n} \alpha_r=S(1).r=0∑n​αr​=S(1).

Now evaluate:

S(1)=∑k=0n−1(1+3)n−1−k(1+2)k=∑k=0n−14n−1−k3k.S(1)=\sum_{k=0}^{n-1}(1+3)^{n-1-k}(1+2)^k =\sum_{k=0}^{n-1}4^{n-1-k}3^k.S(1)=k=0∑n−1​(1+3)n−1−k(1+2)k=k=0∑n−1​4n−1−k3k.

Thus,

S(1)=4n−1+4n−23+4n−332+⋯+3n−1.S(1)=4^{n-1}+4^{n-2}3+4^{n-3}3^2+\cdots+3^{n-1}.S(1)=4n−1+4n−23+4n−332+⋯+3n−1.
  1. Recognize it as a geometric-type identity

Use the standard identity

an−1+an−2b+an−3b2+⋯+bn−1=an−bna−b(a≠b).a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+b^{n-1}=\frac{a^n-b^n}{a-b} \quad (a\ne b).an−1+an−2b+an−3b2+⋯+bn−1=a−ban−bn​(a=b).

With a=4a=4a=4 and b=3b=3b=3,

S(1)=4n−3n4−3=4n−3n.S(1)=\frac{4^n-3^n}{4-3}=4^n-3^n.S(1)=4−34n−3n​=4n−3n.

Hence,

∑r=0nαr=4n−3n.\sum_{r=0}^{n} \alpha_r=4^n-3^n.r=0∑n​αr​=4n−3n.

Comparing with

βn−γn,\beta^n-\gamma^n,βn−γn,

we get

β=4,γ=3.\beta=4,\qquad \gamma=3.β=4,γ=3.
  1. Compute the required value
β2+γ2=42+32=16+9=25.\beta^2+\gamma^2=4^2+3^2=16+9=25.β2+γ2=42+32=16+9=25.
  1. Final answer
25\boxed{25}25​

The derived answer matches the stored correct answer.

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