Prove by mathematical induction that `sum_(r=0)^(n)r^(n)C_(r)=n.2^(n-1), forall n in N`.

Explanation

Let `P(n):sum_(r=0)^(n)r^(n)C_(r)=n.2^(n-1)`
Step I For n=1,
LHS of Eq. (i) `=sum_(r=0)^(1)r.^(1)C_(r)=0+1.^(1)C_(1)=1` and RHS of Eq. (i) `=1.2^(1-1)=2^(0)=1`
Therefore , P(1) is true .
Step II Assume that P(k) is true , then P(k) : `sum_(r=0)^(k)r.^(k)C_(r)=k.2^(k-1)`
Step III For `n=k+1`
`P(k+1):sum_(r=0)^(k+1)r.^(k+1)C_(r)=(k+1).2^(k)`
`therefore LHS =sum_(r=0)^(k+1)r.^(k+1)C_(r)=0+sum_(r=1)^(k+1)r.^(k+1)C_(r)`
`=sum_(r=1)^(k+1)r.^(k+1)C_(r)=sum_9r=1)^(k)r.^(k+1)C_(r)+(k+1).^(k+1)C_(k+1)`
`=sum_(r=1)^(k)r(.^(k)C_(r)+.^(k)C_(r-1))+(k+1)`
`=sum_(r=1)^(k)r.^(k)C_(r)+sum_(r=0)^(k)r.^(k)C_(r-1)+(k+1)`
`=sum_(r=0)^(k)r.^(k)C_(r)+sum_(r=0)^(k+1)r.^(k)C_(r-1)`
`=sum_(r=0)^(k)r.^(k)C_(r)+sum _(r=0)^(k)(r+1).^(k)C_r)`
`=sum_(r=0)^(k)r.^(k)C_(r)+sum_(r=0)^(k)r.^(k)C_(r)+sum_(r=0)^(k).^(k)C_(r)`
`=P(k)+P(k)+2^k` [by assumption step]
`=k.2^(k-1)+k.26(k-1)+2^(k)=2.k.2^k-1+2^k`
`=k.2^k+2^k=(k+1).2^k=RHS`
Therefore , `P(k+1)` is true. Hence , by the principle of mathematical induction `P(n)` is true for all `n in N`.