Question

If a and d are two complex numbers, then the sum

to (n+1) terms of the following series

a$C_0$ - (a + d)$C_1$ + (a + 2d)$C_2$ - ........... is

• A.
• B. na
• C. 0
• D. None of these

Answer 1

The correct option is C

0

We can write

a$C_0$ - (a + d)$C_1$ + (a + 2d)$C_2$ - ........ upto (n+1) terms

$=a(C_0-C_1+C_2-........)+d(-C_1+2C_2-3C_3+......)$ .............(i)

Again,$(1-x{)}^n=C_0-C_{1}x+C_2{x}^2-.........+(-1{)}^n{C}_n{x}^n$ ..........(ii)

Differentiating with respect to x,

$-n(1-x{)}^{n-1}$ = -$C_1$ + 2$C_2$x - .......... + $(-1{)}^n{C}_{n}n{x}^{n-1}$ ......................(iii)

Putting x = 1 in (ii) and (iii), we get

$C_0$ - $C_1$ + $C_2$ - ........ + (-1)${}^n{C}_n$ = 0

and -$C_1$ + 2$C_2$ - ........+$(-1{)}^{n}n.C_n$ = 0

Thus the required sum to (n+1) terms, by (i)

= a.0 + d.0 = 0.