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

Asked by Abhisek | 1 year ago |  74

##### Solution :-

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

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