Question

The sum of coefficient in (x+y)^10 is-
a)1
b)10
c) 512
d) 1024​

Answer 1

Since the terms in this forms do not have any constants in them to get the sum of coefficients of all the expanded terms we just have to add the "C" values,

Let's do it,

10C0+ 10C 1 + 10C2 +10C3 +10C4 +10C5 + 10C6+10C7+10C8+ 10C9+ 10C10

which on furthur solving gives the following values,

$\frac{10!}{(10 - 0) !(0!)} + \frac{10!}{(9!)(1!)} + \frac{10!}{(8) !(2!)} \\ + \frac{10!}{(7) !(3!)} + \frac{10!}{(6) !(4!)} + \frac{10!}{(5) !(5!)} + \frac{10!}{(4) !(6!)} + \\\frac{10!}{(3) !(7!)} + \frac{10!}{(2) !(8!)} + \frac{10!}{(1) !(9!)} + \frac{10!}{(10) !(0!)}$

Which equates to,

$2( \frac{10!}{(10 ) !(0!)} + \frac{10!}{(9!)(1!)} + \frac{10!}{(8) !(2!)} \\ + \frac{10!}{(7) !(3!)} + \frac{10!}{(6) !(4!)} ) + \frac{10!}{(5) !(5!)}\\ \\ = 2(1 + 10 + 45 + 120 + 210 ) + 252 \\ \\ = 1024$

Hence option d is correct i.e 1024.



Huh! Took time but was worth it.