- Home
- Standard 11
- Mathematics
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
$\frac{a}{{{2^n}}}$
$na$
$0$
None of these
Solution
(c) 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} + 2{C_2} – 3{C_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 $(i) =a.0 + d.0 = 0.$
