# Part B) only Show that for integers k and n such that `1<=k<=n` , A) `k^nC_k = n ^(n-1)C_(k-1)` B) Hence or otherwise prove that for any `x in RR` and n>=0, `sum_(k=0)^n k^nC_kx^k (1-x)^(n-k) = nx`

we know that;

`k(^nC_k) = n(^(n-1)C_(k-1))`

`sum_(k=0)^n k (^nC_k) x^k (1-x)^(n-k)`

`=sum_(k=1)^n k (^nC_k) x^k (1-x)^(n-k)`  When k = 0 first term is 0.

`=sum_(k=1)^n n[^(n-1)C_(k-1)] x^k (1-x)^(n-k)`

`=nxsum_(k=1)^n [^(n-1)C_(k-1)] x^(k-1) (1-x)^((n-1)-(k-1))`

let k-1 = j

Since at start k = 1 then at start j = 1-1 = 0

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we know that;

`k(^nC_k) = n(^(n-1)C_(k-1))`

`sum_(k=0)^n k (^nC_k) x^k (1-x)^(n-k)`

`=sum_(k=1)^n k (^nC_k) x^k (1-x)^(n-k)`  When k = 0 first term is 0.

`=sum_(k=1)^n n[^(n-1)C_(k-1)] x^k (1-x)^(n-k)`

`=nxsum_(k=1)^n [^(n-1)C_(k-1)] x^(k-1) (1-x)^((n-1)-(k-1))`

let k-1 = j

Since at start k = 1 then at start j = 1-1 = 0

`=nxsum_(j=0)^(n-1) [^(n-1)C_j] x^j (1-x)^(n-j)`

Using binomial theorem we know that;

`(a+x)^n = sum_(r = 1)^n^nC_ra^rx^(n-r)`

Similarly;

`=nxsum_(j=0)^(n-1) [^(n-1)C_j] x^j (1-x)^(n-j)`

`=nx[(x+(1-x))^(n-1)]`

`=nx`

So the answer is obtained as required.