IIT JEE , AIEEE Forum - Mathematics , Algebra - The coefficient of x24 in ( 1 + 3x + 6x2 + 10x3 + .....)2/3

chaitu_98103

a-25 b-125 c-50 d-300

sriram.a

Frombinomial theorem

$(1-x)^{-n}=1+nx+n(n+1)/2+n(n+1)(n+3)/3+........$

substitute n=3in above equation

$(1-x)^{-3}=1+3x+6x^2+10x^3+.......$

( 1 + 3x + 6x2 + 10x3 + .....)2/3= $(1-x)^{-3\frac{2}{3}}=(1-x)^{-2}$

=1+2x+3x^2+4x^3+....+25x^24+......

so,x^24 coefficient is25(a)

Hope u got it cheeeeeeeeeerrrrrrrrrrsssssss.

iitkgp_bipin

Good work sriram.

The expression inside the bracket is expansion of (1-x)^-3.

so the expression becomes {(1-x)^-3}^2/3 = (1-x)^-2 which on expansion gives 25 as the coefficient of x²⁴.