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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
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
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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