If ∑_ r=1 ^ 30 r^2( ^ 30 C_r)^2 ^ 30 C_ r-1 =α × 2^ 29 , then α is equal to ______

Correct answer: 0 aligned & ∑_ r=1 ^ 30 r^2( ^ 30 C_ r )^2 ^ 30 C_ r-1 & =∑_ r=1 ^ 30 r^2( 31-r r ) · 30! r!(30-r)! & ( ^ 30 C_ r ^ 30 C_ r-1 = 30-r+1 r = 31-r r ) aligned aligned & =∑_ r=1 ^ 30 (31-r) 30! (r-1)!(30-r)! & =30 ∑_ r=1 ^ 30 (31-r) 29! (r-1)!(30-r)! & =30 ∑_ r=1 ^ 30 (30-r+1)^ 29 C_ 30-r & =30(∑_ r=1 ^ 30 (31-r)^ 29 C_ 30-r +∑_ r=1 ^ 30 ^ 29 C_ 30-r ) & =30(29 × 2^ 28 +2^ 29 )=30(29+2) 2^ 28 & =15 × 31 × 2^ 29 & =465(2^ 29 ) & α=465 aligned

JEE Main 202522 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Binomial Theorem

If ∑_ r=1 ^ 30 r^2( ^ 30 C_r)^2 ^ 30 C_ r-1 =α × 2^ 29 , then α is...

\sum_{r=1}^{30} \frac{r^2\left({ }^{30} C_r\right)^2}{{ }^{30} C_{r-1}}=\alpha \times 2^{29}

, then

\alpha

is equal to ______

Correct answer

Step-by-step solution

\begin{aligned} & \sum_{\mathrm{r}=1}^{30} \frac{\mathrm{r}^2\left({ }^{30} \mathrm{C}_{\mathrm{r}}\right)^2}{{ }^{30} \mathrm{C}_{\mathrm{r}-1}} \\ & =\sum_{\mathrm{r}=1}^{30} \mathrm{r}^2\left(\frac{31-\mathrm{r}}{\mathrm{r}}\right) \cdot \frac{30!}{\mathrm{r}!(30-\mathrm{r})!} \\ & \left(\because \frac{{ }^{30} \mathrm{C}_{\mathrm{r}}}{{ }^{30} \mathrm{C}_{\mathrm{r}-1}}=\frac{30-\mathrm{r}+1}{\mathrm{r}}=\frac{31-\mathrm{r}}{\mathrm{r}}\right)\end{aligned}

\begin{aligned} & =\sum_{\mathrm{r}=1}^{30} \frac{(31-\mathrm{r}) 30!}{(\mathrm{r}-1)!(30-\mathrm{r})!} \\ & =30 \sum_{\mathrm{r}=1}^{30} \frac{(31-\mathrm{r}) 29!}{(\mathrm{r}-1)!(30-\mathrm{r})!} \\ & =30 \sum_{\mathrm{r}=1}^{30}(30-\mathrm{r}+1)^{29} \mathrm{C}_{30-\mathrm{r}} \\ & =30\left(\sum_{\mathrm{r}=1}^{30}(31-\mathrm{r})^{29} \mathrm{C}_{30-\mathrm{r}}+\sum_{\mathrm{r}=1}^{30}{ }^{29} \mathrm{C}_{30-\mathrm{r}}\right) \\ & =30\left(29 \times 2^{28}+2^{29}\right)=30(29+2) 2^{28} \\ & =15 \times 31 \times 2^{29} \\ & =465\left(2^{29}\right) \\ & \alpha=465\end{aligned}

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