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JEE Mathematics 2025 Question with Solution

If ∑r=19(r+32r)⋅9Cr=α(32)9−β\sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r} = \alpha \left( \frac{3}{2} \right)^9 - \beta, α,β∈N\alpha, \beta \in \mathbb{N}, then (α+β)2(\alpha + \beta)^2 is equal to

  • A

    2727

  • B

    99

  • C

    8181

  • D

    1818

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

∑r=19(r+32r)⋅9Cr=α(32)9−β,α,β∈N\sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r} = \alpha \left( \frac{3}{2} \right)^9 - \beta, \quad \alpha, \beta \in \mathbb{N}

Find: (α+β)2(\alpha + \beta)^2

Split the summation into two parts:

∑r=19(r+32r)⋅9Cr=∑r=19(r2r⋅9Cr)+∑r=19(32r⋅9Cr)\sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r} = \sum_{r=1}^{9} \left( \frac{r}{2^r} \cdot {^9C_r} \right) + \sum_{r=1}^{9} \left( \frac{3}{2^r} \cdot {^9C_r} \right)

Use the identity

r⋅nCr=n⋅n−1Cr−1r \cdot {^nC_r} = n \cdot {^{n-1}C_{r-1}}

So,

r2r⋅9Cr=92r⋅8Cr−1\frac{r}{2^r} \cdot {^9C_r} = \frac{9}{2^r} \cdot {^8C_{r-1}}

Hence,

∑r=19r2r⋅9Cr=9∑r=1912r⋅8Cr−1=92∑r=198Cr−1⋅(12)r−1\sum_{r=1}^{9} \frac{r}{2^r} \cdot {^9C_r} = 9 \sum_{r=1}^{9} \frac{1}{2^r} \cdot {^8C_{r-1}} = \frac{9}{2} \sum_{r=1}^{9} {^8C_{r-1}} \cdot \left( \frac{1}{2} \right)^{r-1}

Put s=r−1s = r-1. Then ss goes from 00 to 88:

92∑s=088Cs(12)s=92⋅(1+12)8=92⋅(32)8\frac{9}{2} \sum_{s=0}^{8} {^8C_s} \left( \frac{1}{2} \right)^s = \frac{9}{2} \cdot \left(1 + \frac{1}{2}\right)^8 = \frac{9}{2} \cdot \left( \frac{3}{2} \right)^8

For the second part,

∑r=1932r⋅9Cr=3(∑r=099Cr⋅(12)r−9C0⋅1)=3((1+12)9−1)=3((32)9−1)\sum_{r=1}^{9} \frac{3}{2^r} \cdot {^9C_r} = 3 \left( \sum_{r=0}^{9} {^9C_r} \cdot \left( \frac{1}{2} \right)^r - {^9C_0} \cdot 1 \right) = 3 \left( \left( 1 + \frac{1}{2} \right)^9 - 1 \right) = 3 \left( \left( \frac{3}{2} \right)^9 - 1 \right)

Add both parts:

92⋅(32)8+3((32)9−1)=(92⋅23+3)⋅(32)9−3=(3+3)⋅(32)9−3=6(32)9−3\frac{9}{2} \cdot \left( \frac{3}{2} \right)^8 + 3 \left( \left( \frac{3}{2} \right)^9 - 1 \right) = \left( \frac{9}{2} \cdot \frac{2}{3} + 3 \right) \cdot \left( \frac{3}{2} \right)^9 - 3 = (3 + 3) \cdot \left( \frac{3}{2} \right)^9 - 3 = 6 \left( \frac{3}{2} \right)^9 - 3

Comparing with

α(32)9−β\alpha \left( \frac{3}{2} \right)^9 - \beta

we get α=6\alpha = 6 and β=3\beta = 3.

Therefore,

(α+β)2=(6+3)2=81(\alpha + \beta)^2 = (6 + 3)^2 = 81

The correct option is C.

Detailed Binomial Expansion

Given:

S=∑r=19(r+32r)⋅9CrS = \sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r}

and

S=α(32)9−βS = \alpha \left( \frac{3}{2} \right)^9 - \beta

Find: (α+β)2(\alpha + \beta)^2

Let

S1=∑r=19r9Cr2r,S2=∑r=1939Cr2rS_1 = \sum_{r=1}^{9} r \frac{{^9C_r}}{2^r}, \quad S_2 = \sum_{r=1}^{9} 3 \frac{{^9C_r}}{2^r}

Then

S=S1+S2S = S_1 + S_2

Using

r⋅9Cr=9⋅8Cr−1r \cdot {^9C_r} = 9 \cdot {^8C_{r-1}}

we get

S1=∑r=199⋅8Cr−12r=9∑r=198Cr−12⋅2r−1=92∑r=198Cr−1(12)r−1S_1 = \sum_{r=1}^{9} \frac{9 \cdot {^8C_{r-1}}}{2^r} = 9 \sum_{r=1}^{9} \frac{{^8C_{r-1}}}{2 \cdot 2^{r-1}} = \frac{9}{2} \sum_{r=1}^{9} {^8C_{r-1}} \left(\frac{1}{2}\right)^{r-1}

Now let k=r−1k = r-1. Then

S1=92∑k=088Ck(12)kS_1 = \frac{9}{2} \sum_{k=0}^{8} {^8C_k} \left(\frac{1}{2}\right)^k

By the binomial theorem,

∑k=088Ck(12)k=(1+12)8=(32)8\sum_{k=0}^{8} {^8C_k} \left(\frac{1}{2}\right)^k = \left(1 + \frac{1}{2}\right)^8 = \left(\frac{3}{2}\right)^8

Hence,

S1=92(32)8=3(32)9S_1 = \frac{9}{2} \left(\frac{3}{2}\right)^8 = 3 \left(\frac{3}{2}\right)^9

Next,

S2=3∑r=199Cr(12)rS_2 = 3 \sum_{r=1}^{9} {^9C_r} \left(\frac{1}{2}\right)^r

Now,

∑r=099Cr(12)r=(1+12)9=(32)9\sum_{r=0}^{9} {^9C_r} \left(\frac{1}{2}\right)^r = \left(1 + \frac{1}{2}\right)^9 = \left(\frac{3}{2}\right)^9

Therefore,

∑r=199Cr(12)r=(32)9−1\sum_{r=1}^{9} {^9C_r} \left(\frac{1}{2}\right)^r = \left(\frac{3}{2}\right)^9 - 1

So,

S2=3(32)9−3S_2 = 3 \left(\frac{3}{2}\right)^9 - 3

Thus,

S=S1+S2=3(32)9+3(32)9−3=6(32)9−3S = S_1 + S_2 = 3 \left(\frac{3}{2}\right)^9 + 3 \left(\frac{3}{2}\right)^9 - 3 = 6 \left(\frac{3}{2}\right)^9 - 3

Comparing with the required form,

α=6,β=3\alpha = 6, \quad \beta = 3

Finally,

α+β=9\alpha + \beta = 9

and

(α+β)2=92=81(\alpha + \beta)^2 = 9^2 = 81

Therefore, the value is 8181, so the correct option is C.

Common mistakes

  • A common mistake is to apply the binomial theorem directly to ∑r=19r+32r9Cr\sum_{r=1}^{9} \frac{r+3}{2^r}{^9C_r} without first splitting the term r+3r+3. This is wrong because the factor rr requires the identity r nCr=n n−1Cr−1r\,{^nC_r} = n\,{^{n-1}C_{r-1}}. Split the sum into two separate sums before evaluating.

  • Students often forget that the second sum starts from r=1r=1, not r=0r=0. This is wrong because the full binomial sum includes the r=0r=0 term equal to 11. After using ∑r=099Cr(12)r=(32)9\sum_{r=0}^{9} {^9C_r}\left(\frac{1}{2}\right)^r = \left(\frac{3}{2}\right)^9, subtract the missing r=0r=0 term.

  • Another mistake is in changing variables from rr to r−1r-1 and keeping incorrect limits. This leads to a wrong binomial sum. When you substitute s=r−1s = r-1, the limits change from r=1r=1 to 99 into s=0s=0 to 88.

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