7.Binomial Theorem
hard

જો $\frac{1}{n+1}{ }^n C_n+\frac{1}{n}{ }^n C_{n-1}+\ldots+\frac{1}{2}{ }^{ n } C _1+{ }^{ n } C _0=\frac{1023}{10}$ હોય,તો $n=..........$

A

$6$

B

$9$

C

$8$

D

$7$

(JEE MAIN-2023)

Solution

$\sum \limits_{ r =0}^{ n } \frac{{ }^{ n } C_{ r }}{ r +1}=\frac{1}{ n +1} \sum \limits_{ r =0}^{ n }{ }^{ n +1} C_{ r +1}$

$=\frac{1}{ n +1}\left(2^{ n +1}-1\right)=\frac{1023}{10}$

$n +1=10 \Rightarrow n =9$

Standard 11
Mathematics

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