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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
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,
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, 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
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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