1
Manmay kumar Mohanty
·Apr 4 '10 at 21:34
2)
for this i tried like this
(1-x)n = C0 - C1x + C2x2............
now multiplying both the sides by xn-1 and integrate with limits o to 1.
u will get
∫01xn−1(1−x)ndx=nC0−n+1C1+n+2C2............
to integrate L.H.S substitute x = sin2θ Integral reduces to 2∫0π/2sin2n−2θcos2n+1θdθ
u will need to use gamma function which is as ∫0π/2sinmxcosnxdx=2Γ(2m+n+2)Γ(2m+1)Γ(2n+1),m,nϵR
and Γn=(n−1)! if nϵN hence integral reduces to
2∫0π/2sin2n−2θcos2n+1θdθ = 2Γ(22n−1+2n+1+2)2.Γ(22n−1+1)Γ(22n+1+1)=Γ(2n+1)ΓnΓ(n+1)=2n!(n−1)!n!
wat to do next can't thnk
[7][7]
11
SANDIPAN CHAKRABORTY
·Apr 5 '10 at 7:00
for Q1 i'm half done..
∑r=0n−1(n−r)21(CrCr+1)2
∑r=0n−1(n−r)21(1+rn−(1+r)+1)2
∑r=0n−1(n−r)21(1+rn−r)2
∑r=0n−1(r+1)21
after this i got stuck [12][12][12][12]
1
Manmay kumar Mohanty
·Apr 5 '10 at 7:14
Q1) CrCr+1=(n−r)!r!n!(n−(r−1))!(r+1)!n!=(n−(r−1))!(n−r)!=(n−r)
so ∑r=0n−1(n−r)21(CrCr+1)2 = ∑r=0n−1(n−r)21(n−r)2/(r+1)
same as sandipan
= ∑r=0n−1r+11
now take summation of the series
that's wat i got ??
edited
11
SANDIPAN CHAKRABORTY
·Apr 5 '10 at 7:42
nCr+1nCr
= n!(n-(r+1))! (r+1)!n!(n-r)! r!
= (n-r)! r! (n-(r+1))! (r+1)!
= (n-r) (n-r-1)! r!(n-(r+1)! (r+1) r!
= (n-r) (n-r-1)! r!(n-r-1)! (r+1) r!
=n - rr + 1
11
SANDIPAN CHAKRABORTY
·Apr 5 '10 at 10:53
@manmay you missed the square.... it will be summation 1/(1+r)2......
1
rvking
·Apr 5 '10 at 11:41
conti....frm sandipan's half solution:
\sum_{r=0}^{n-1}{[1/r+1]^{2}}
=1/1+1/4+1/9+1/16......1/n^2
=\Pi ^{2/6}
11
SANDIPAN CHAKRABORTY
·Apr 5 '10 at 12:11
@rvking i could'nt get what you are trying to tell....
how 11+122+132+....+1n2 = π2/6
please explain
1
rvking
·Apr 5 '10 at 12:58
i cant xplain the solution of this series.....its something we must rote learn...
cause the derivation is too complicated....though u can chk on the link given by Carl.
62
Lokesh Verma
·Apr 5 '10 at 22:43
actually it is not π2/6 but π2/6 and as you all said this is a fact which we should not worry beyond a point at this time... just a fact.