7.Binomial Theorem
hard

यदि $C _{ x } \equiv{ }^{25} C _{ x }$ तथा $C _{0}+5 \cdot C _{1}+9 \cdot C _{2}+\ldots+$ (101). $C _{25}=2^{25} \cdot k$, तो $k$ बराबर है

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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