If ∑_ r=0 ^ 10 ( 10^ r+1 -1 10^ r ) · ^ 11 C_ r+1 = α^ 11 -11^ 11 10^ 10 , then α is equal to :

Correct answer: D. 20 aligned & ∑_ r=0 ^ 10 ( 1 r^ r-1 -1 10^ r ) ^ 11 C_ r+1 & =∑_ r=0 ^ 10 (10- 1 10^ r )^ 11 C_ r+1 & =10 ∑_ r=0 ^ 10 C_ r+1 -10 ∑( ^ 11 C_ r+1 ((1)/(10))^ r+1 ) & =10[ ^ 11 C_1+ ^ 11 C_2+ . .+ ^ 11 C_ 11 ] & -10[ ^ 11 C_1((1)/(10))^1+ ^ 11 C_2((1)/(10))^2+ . .+ ^ 11 C_ 11 ((1)/(10))^ 11 ] & =10[2^ 11 -1]-10[(1+(1)/(10))^ 11 -1] & =10(2)^ 11 -10- 11^ 11 10^ 10 +10 & = (20)^ 11 -11^ 11 10^ 10 & α=20 aligned

JEE Main 20252 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Binomial Theorem

If ∑_ r=0 ^ 10 ( 10^ r+1 -1 10^ r ) · ^ 11 C_ r+1 = α^ 11 -11^ 11...

\sum_{\mathrm{r}=0}^{10}\left(\frac{10^{\mathrm{r}+1}-1}{10^{\mathrm{r}}}\right) \cdot{ }^{11} \mathrm{C}_{\mathrm{r}+1}=\frac{\alpha^{11}-11^{11}}{10^{10}}

, then

\alpha

is equal to :

Options

A $15$
B $11$
C $24$
D $20$

Correct answer

D. $20$

Step-by-step solution

\begin{aligned} & \sum_{\mathrm{r}=0}^{10}\left(\frac{1 \mathrm{r}^{\mathrm{r}-1}-1}{10^{\mathrm{r}}}\right){ }^{11} \mathrm{C}_{\mathrm{r}+1} \\ & =\sum_{\mathrm{r}=0}^{10}\left(10-\frac{1}{10^{\mathrm{r}}}\right)^{11} \mathrm{C}_{\mathrm{r}+1} \\ & =10 \sum_{\mathrm{r}=0}^{10} \mathrm{C}_{\mathrm{r}+1}-10 \sum\left({ }^{11} \mathrm{C}_{\mathrm{r}+1}\left(\frac{1}{10}\right)^{\mathrm{r}+1}\right) \\ & =10\left[{ }^{11} \mathrm{C}_1+{ }^{11} \mathrm{C}_2+\ldots . .+{ }^{11} \mathrm{C}_{11}\right] \\ & -10\left[{ }^{11} \mathrm{C}_1\left(\frac{1}{10}\right)^1+{ }^{11} \mathrm{C}_2\left(\frac{1}{10}\right)^2+\ldots . .+{ }^{11} \mathrm{C}_{11}\left(\frac{1}{10}\right)^{11}\right] \\ & =10\left[2^{11}-1\right]-10\left[\left(1+\frac{1}{10}\right)^{11}-1\right] \\ & =10(2)^{11}-10-\frac{11^{11}}{10^{10}}+10 \\ & =\frac{(20)^{11}-11^{11}}{10^{10}} \\ & \therefore \alpha=20\end{aligned}

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