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Prove that sum(r=0)^n3^r^n Cr=4^n....

Prove that `sum_(r=0)^n3^r^n C_r=4^n`.

Text Solution

Verified by Experts

By Binomial Theorem,

` " " " " " ` `sum_(r=0)^n " ^n C_r a^(n-r) b^r=(a+b)^n`

Putting `b=3` and `a=1` in the above equation

` " " " " sum_(r=0)^n " ^n C_r 1^(n-r) 3^r=(1+3)^n`

` " " " " sum_(r=0)^n 3^r " " ^n C_r =(4)^n`

Hence Proved
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