Prove that \displaystyle\sum_{r=0}^{n} 3^r \;^nC_r=4^n

Question

Prove that \(\displaystyle\sum_{r=0}^{n} 3^r \;^nC_r=4^n\)

Asked by Abhisek | 1 year ago | 74

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Accepted Answer

By Binomial Theorem,

\( \displaystyle\sum_{r=0}^{n}\; ^nC_ra^{n-r}b^r=(a+n)^n\)

By putting b =3 and a =1 in the above equation, we obtain

\( \displaystyle\sum_{r=0}^{n}\; ^nC_r(1)^{n-r}(3)^r=(1+3)^n\)

\( \displaystyle\sum_{r=0}^{n}3^r\;^nC_r=4^n\)

Hence proved.

Answered by Pragya Singh | 1 year ago