Question

If ( 1 + x ) n = c 0 + c 1 x + c 2 x 2 + ⋯ + c n x n then show that 3 . n C 0 − 8 . n C 1 + 13 . n C 2 − 18 . n C 3 + ... . upto ( n + 1 ) terms = 0 ?

Answer 1

See below.

Explanation:

Assuming that the question reads

If `( 1 + x )^n = c_0 + c_1 x + c_2 x^2 + ⋯ + c_n x^n` then show that `3 . n c_0 − 8 . n c_1 + 13 . n c_2 − 18 . n c_3 + cdots + = 0` ?

This is equivalent to

`n(sum_(k=0)^n (-1)^k(3+5k)c_k)=3n sum_(k=0)^n(-1)^k c_k + 5n sum_(k=0)^n(-1)^k k c_k`

but `sum_(k=0)^n(-1)^k c_k =0`

because it is obtained from

`(1+x)^n = sum_(k=0)^n c_k x^k` by making `x=-1`

and `sum_(k=0)^n(-1)^k k c_k = 0`

because is is obtained from

`d/(dx)(1-x)^n = sum_(k=0)^n k c_k x^(k-1)` by making `x = -1`

then

`n(sum_(k=0)^n (-1)^k(3+5k)c_k)=0`