If C_{x} equiv^{25} C_{x} and mathrm{C}_{0}+5 · mathrm{C}_{1}+9 · mathrm{C}_{2}+ldots .+(101) · m

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7.Binomial Theorem

hard

If $C_{x} \equiv^{25} C_{x}$ and $\mathrm{C}_{0}+5 \cdot \mathrm{C}_{1}+9 \cdot \mathrm{C}_{2}+\ldots .+(101) \cdot \mathrm{C}_{25}=2^{25} \cdot \mathrm{k}$ then $\mathrm{k}$ is equal to

A

$42$

B

$45$

C

$51$

D

$48$

(JEE MAIN-2020)

Solution

$\mathrm{S}=1 .^{25} \mathrm{C}_{0}+5.2^{25} \mathrm{C}_{1}+9.2^{25} \mathrm{C}_{2}+\ldots .+(101)^{25} \mathrm{C}_{25}$

$\mathrm{S}=101^{25} \mathrm{C}_{25}+97^{25} \mathrm{C}_{1}+\ldots \ldots \ldots .+1^{25} \mathrm{C}_{25}$

$2 \mathrm{S}=(102)\left(2^{25}\right)$

$\mathrm{S}=51\left(2^{25}\right)$

Standard 11

Mathematics

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(JEE MAIN-2024)