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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