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```        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
7 years ago

28 Points
```										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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All the best H V !!!

Regards,
```
7 years ago
h v
33 Points
```										can anyone give the answer please?
```
6 years ago
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