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If (1+x) n =C 0 +C 1 x+C 2 x 2 ....+C n x n ,then C 0 +3C 1 +5C 2 +...+(2n+1)C n is equal to (a)n*2 n (b)(n+1)2 n (c)n*2 n +1 (d)0

If (1+x)n=C0+C1x+C2x2....+Cnxn,then


C0+3C1+5C2+...+(2n+1)Cn  is equal to


(a)n*2n


(b)(n+1)2n


(c)n*2n+1


(d)0

Grade:12

2 Answers

askiitian expert akshay singh
8 Points
13 years ago

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.

Pratham Ashish
17 Points
13 years ago

hi vaibhav,

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

Tr = (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

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