# dear sir,solve for n;(2n+1)C1 here is:out of 2n+1 objects, 1 object selected...and not (2n+1)*C1(2n+1)C1 + (2n+1)C2 + (2n+1)C3 +....+(2n+1)Cn = 63[answer ; n=3]

28 Points
14 years ago

Dear H V,

Ans:- Let us consider the expansion (1+X)^(2n+1)

(1+X)^(2n+1)=(2n+1)C0+(2n+1)C1 X +(2N+1)C2 X²+............+(2N+1)C2N X^2N+(2N+1)C(2N+1)X^2N+1................1

Now we know that NCr=NC(N-r)

(2N+1)C1=(2n+1)C2N

(2N+1)C2=(2N+1)C(2N-1)

Again ,

(2N+1)CN=(2N+1)C(N+1) etc

Hence from these relations we get,(Putting X=1 in eq 1)

2^(2N+1)=(2N+1)C0+2K+(2N+1)C(2N+1)

where K=(2N+1)C1+(2N+1)C2+..............(2N+1)C(2N+1)=63(given)

2^(2N+1)=1+1+126=128

4^N=64=4³

Hence N=3(Ans)

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Soumyajit Das IIT Kharagpur

h v
33 Points
13 years ago

sunil
13 Points
6 years ago
formula:
(2n+1)C0 +(2n+1)C1+ (2n+1)C2 + (2n+1)C3 +....+(2n+1)Cn=1/2{(2n+1)C0 +(2n+1)C1+ (2n+1)C2 + (2n+1)C3 +....+(2n+1)C2n+1}
again (2n+1)C0 + (2n+1)C2 + (2n+1)C3 +....+(2n+1)C2n+1= 2^(2n+1)
so (2n+1)C0 +(2n+1)C1+ (2n+1)C2 + (2n+1)C3 +....+(2n+1)Cn=1/2(2^(2n+1))....eq 1
given:
(2n+1)C1 + (2n+1)C2 + (2n+1)C3 +....+(2n+1)Cn = 63
put this in eq1
so (2n+1)C0 +63=1/2(2^(2n+1))
64=1/2(2^(2n+1)) as (2n+1)C0=1
solve this n comes out to be 3