Logo
Jeeify
MCQMediumJEE 2025Binomial Expansion

JEE Mathematics 2025 Question with Solution

If ∑r=010(10r+1−1)(10r)=α11−1\sum_{r=0}^{10} \left( 10^{r+1} - 1 \right) \binom{10}{r} = \alpha^{11} - 1 then α\alpha is equal to :

  • A

    1515

  • B

    1111

  • C

    2424

  • D

    2020

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

∑r=010(10r+1−110r)⋅11Cr+1=α11−11111010\sum_{r=0}^{10} \left( \frac{10^{r+1}-1}{10^r} \right) \cdot {}^{11}C_{r+1} = \frac{\alpha^{11} - 11^{11}}{10^{10}}

Find: α\alpha

Let the left-hand side be SS. First simplify the term inside the summation:

10r+1−110r=10r+110r−110r=10−110r\frac{10^{r+1}-1}{10^r} = \frac{10^{r+1}}{10^r} - \frac{1}{10^r} = 10 - \frac{1}{10^r}

Now change the index by taking k=r+1k = r+1. Then as rr goes from 00 to 1010, kk goes from 11 to 1111:

S=∑r=010(10−110r)11Cr+1=∑k=111(10−110k−1)11CkS = \sum_{r=0}^{10} \left( 10 - \frac{1}{10^r} \right) {}^{11}C_{r+1} = \sum_{k=1}^{11} \left( 10 - \frac{1}{10^{k-1}} \right) {}^{11}C_k

Split the summation into two parts:

S=∑k=11110⋅11Ck−∑k=111110k−1⋅11CkS = \sum_{k=1}^{11} 10 \cdot {}^{11}C_k - \sum_{k=1}^{11} \frac{1}{10^{k-1}} \cdot {}^{11}C_k S=10∑k=11111Ck−10∑k=11111Ck(110)kS = 10 \sum_{k=1}^{11} {}^{11}C_k - 10 \sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k

Use the identity for the sum of binomial coefficients:

∑k=11111Ck=211−1\sum_{k=1}^{11} {}^{11}C_k = 2^{11} - 1

So the first part becomes 10(211−1)10(2^{11}-1).

For the second part, use the binomial expansion of (1+110)11\left(1+\frac{1}{10}\right)^{11}:

(1+110)11=∑k=01111Ck(110)k\left(1 + \frac{1}{10}\right)^{11} = \sum_{k=0}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k (1110)11=1+∑k=11111Ck(110)k\left(\frac{11}{10}\right)^{11} = 1 + \sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k ∑k=11111Ck(110)k=11111011−1\sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k = \frac{11^{11}}{10^{11}} - 1

Hence the second part becomes

10(11111011−1)=11111010−1010 \left( \frac{11^{11}}{10^{11}} - 1 \right) = \frac{11^{11}}{10^{10}} - 10

Therefore,

S=10(211−1)−(11111010−10)S = 10(2^{11} - 1) - \left( \frac{11^{11}}{10^{10}} - 10 \right) S=10⋅211−11111010S = 10 \cdot 2^{11} - \frac{11^{11}}{10^{10}}

Now equate this with the given right-hand side:

10⋅211−11111010=α11−1111101010 \cdot 2^{11} - \frac{11^{11}}{10^{10}} = \frac{\alpha^{11} - 11^{11}}{10^{10}}

Rewrite the left side with denominator 101010^{10}:

1010⋅(10⋅211)−11111010=α11−11111010\frac{10^{10} \cdot (10 \cdot 2^{11}) - 11^{11}}{10^{10}} = \frac{\alpha^{11} - 11^{11}}{10^{10}} 1011⋅211−11111010=α11−11111010\frac{10^{11} \cdot 2^{11} - 11^{11}}{10^{10}} = \frac{\alpha^{11} - 11^{11}}{10^{10}}

Comparing numerators,

(10⋅2)11−1111=α11−1111(10 \cdot 2)^{11} - 11^{11} = \alpha^{11} - 11^{11} 2011=α1120^{11} = \alpha^{11}

Therefore, the value of α\alpha is 2020. The correct option is D.

Binomial Identity Expansion

Given: the solution rewrites the expression into a form involving 11Cr+1^{11}C_{r+1} and then compares it with

α11−11111010\frac{\alpha^{11} - 11^{11}}{10^{10}}

Find: α\alpha

The main idea is to express the sum in terms of two standard binomial sums:

  1. ∑11Ck\sum {}^{11}C_k
  2. ∑11Ck(110)k\sum {}^{11}C_k \left(\frac{1}{10}\right)^k

After simplification, the sum becomes

S=10∑k=11111Ck−10∑k=11111Ck(110)kS = 10 \sum_{k=1}^{11} {}^{11}C_k - 10 \sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k

Now evaluate each piece separately.

First,

∑k=11111Ck=211−1\sum_{k=1}^{11} {}^{11}C_k = 2^{11} - 1

so the first contribution is

10(211−1)10(2^{11}-1)

Second, from

(1+110)11=1+∑k=11111Ck(110)k\left(1+\frac{1}{10}\right)^{11} = 1 + \sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k

we get

∑k=11111Ck(110)k=11111011−1\sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k = \frac{11^{11}}{10^{11}} - 1

Multiplying by 1010 gives

11111010−10\frac{11^{11}}{10^{10}} - 10

Hence,

S=10(211−1)−(11111010−10)=10⋅211−11111010S = 10(2^{11}-1) - \left(\frac{11^{11}}{10^{10}} - 10\right) = 10\cdot 2^{11} - \frac{11^{11}}{10^{10}}

Comparing this with the right-hand side yields

α11=2011\alpha^{11} = 20^{11}

Therefore, α=20\alpha = 20.

Common mistakes

  • Using ∑k=11111Ck=211\sum_{k=1}^{11} {}^{11}C_k = 2^{11} is incorrect because the term 11C0^{11}C_0 is omitted. Use ∑k=11111Ck=211−1\sum_{k=1}^{11} {}^{11}C_k = 2^{11} - 1 instead.

  • Failing to change the index correctly from rr to k=r+1k=r+1 leads to wrong limits and wrong binomial coefficients. After substitution, the limits must change from r=0 to 10r=0\text{ to }10 into k=1 to 11k=1\text{ to }11.

  • Misreading 110k−1\frac{1}{10^{k-1}} as (110)k−1\left(\frac{1}{10}\right)^{k-1} without carrying the extra factor of 1010 can spoil the second sum. Rewrite it carefully as 1010k\frac{10}{10^k} so that a standard binomial expansion can be applied.

Practice more Binomial Expansion questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Start free trialMore on Binomial Expansion

Related questions