The value of ^nC_1 sum_{r=0}^1 {^1C_r} + ^nC_ | vedclass

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(C) We know that $\sum_{r=0}^k {^kC_r} = 2^k$.Substituting this into the given expression:$= ^nC_1(2^1) + ^nC_2(2^2) + ^nC_3(2^3) + \dots + ^nC_n(2^n)

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$3^n - 1$

Vedclass · Answer

(C) We know that $\sum_{r=0}^k {^kC_r} = 2^k$.Substituting this into the given expression:$= ^nC_1(2^1) + ^nC_2(2^2) + ^nC_3(2^3) + \dots + ^nC_n(2^n)$.Using the Binomial Theorem,$(1+x)^n = \sum_{k=0}^n {^nC_k} x^k = ^nC_0 + ^nC_1 x + ^nC_2 x^2 + \dots + ^nC_n x^n$.For $x=2$,$(1+2)^n = ^nC_0 + ^nC_1(2^1) + ^nC_2(2^2) + \dots + ^nC_n(2^n)$.$3^n = 1 + [^nC_1(2^1) + ^nC_2(2^2) + \dots + ^nC_n(2^n)]$.Therefore,the required sum is $3^n - 1$.

The value of $^nC_1 \sum_{r=0}^1 {^1C_r} + ^nC_2 \left( \sum_{r=0}^2 {^2C_r} \right) + ^nC_3 \left( \sum_{r=0}^3 {^3C_r} \right) + \dots + ^nC_n \left( \sum_{r=0}^n {^nC_r} \right)$ is equal to

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