If (1+x)ⁿ=C₀+C₁x+C₂x²....+C_nxⁿ,then

C₀+3C₁+5C₂+...+(2n+1)C_n is equal to

(a)n*2ⁿ

(b)(n+1)2ⁿ

(c)n*2ⁿ+1

(d)0

hi vaibhav,

  solve this problem using the same trick,put n=2

(1+x)²  =1+2x+x²  ,so c₀=1,c₁=2,c₂=1

c0+3c1+5c2=12,now solve all the options with n=2 u'll find option(b)(n+1)2ⁿ=12 is the right answer.

hi vaibhav,

in dis problem , u can write d general term of d series dat u hav to find,

T_r = (2r + 1)Cr

now apply summation ,

S =2 *  sum r* Cr + sum Cr

sum Cr can be calculated by putting 1 in place of x in (1+x)^n, dat will give 2^n

for r * Cr , u just differentiate (1+x)^n  and again put 1 in place of x , den u will get n*2^(n-1)=C1+2*C2+.......

d final expression after adding will be n*2^n +2^n = (n+1)2^n