✔ 最佳答案
Use binomial expansion, it will be easier to prove then,
we know that
...................n
(1 + x)^n = ∑ C(n,k) x^k
...............k=0
Now diffrentiate wrt "x"
............................n
n*(1 + x)^(n - 1) = ∑ C(n,k) k x^(k - 1)
.........................k=0
In RHS we can start the summation from 1 instead of 0 since fist term will be zero because of the presence of "k", so, we have
............................n
n*(1 + x)^(n - 1) = ∑ C(n,k) k x^(k - 1)
.........................k=1
Now put x = 1 and you will get
......................n
n*(2)^(n - 1) = ∑ k * C(n,k)
...................k=1