q1 S=\sum_{r=0}^{n-1}{\frac{1}{(n-r)^2}}(\frac{C_{r+1}}{C_r})^2
q2 S=\frac{C_0}{n(n+2)}-\frac{C_1}{(n+1)(n+3)}+\frac{C_2}{(n+2)(n+4)}......(n+1) terms
12)
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
\int_{0}^{1}{x^{n-1}(1-x)^{n}}dx = \frac{C_{0}}{n}-\frac{C_{1}}{n+1}+\frac{C_{2}}{n+2}............
to integrate L.H.S substitute x = sin2θ Integral reduces to 2\int_{0}^{\pi /2}{sin^{2n-2}\theta cos^{2n+1}\theta }d\theta
u will need to use gamma function which is as \int_{0}^{\pi /2}{sin^{m}x cos^{n}x }dx= \frac{\Gamma \left(\frac{m+1}{2} \right)\Gamma \left(\frac{n+1}{2} \right)}{2\Gamma \left(\frac{m+n+2}{2} \right)},m,n\epsilon R
and \Gamma n = (n-1)! if n\epsilon N hence integral reduces to
2\int_{0}^{\pi /2}{sin^{2n-2}\theta cos^{2n+1}\theta }d\theta = \frac{2.\Gamma \left(\frac{2n-1+1 }{2}\right)\Gamma \left(\frac{2n+1+1}{2} \right)}{2\Gamma \left(\frac{2n-1+2n+1+2}{2} \right)}=\frac{\Gamma n\Gamma (n+1)}{\Gamma (2n+1)}=\frac{(n-1)!n!}{2n!}
wat to do next can't thnk
[7][7]
11for Q1 i'm half done..
\sum_{r=0}^{n-1}{\frac{1}{(n-r)^{2}}}\left( \frac{C_{r+1}}{C_{r}}\right)^2
\sum_{r=0}^{n-1}{\frac{1}{(n-r)^{2}}}\left( \frac{n-(1+r)+1}{1+r}\right)^2
\sum_{r=0}^{n-1}{\frac{1}{(n-r)^{2}}}\left( \frac{n-r}{1+r}\right)^2
\sum_{r=0}^{n-1}{\frac{1}{(r+1)^{2}}}
after this i got stuck [12][12][12][12]
1Q1) \frac{C_{r+1}}{C_{r}} = \frac{\frac{n!}{(n-(r-1))!(r+1)!}}{\frac{n!}{(n-r)!r!}}= \frac{(n-r)!}{(n-(r-1))!} = (n-r)
so \sum_{r=0}^{n-1}{\frac{1}{(n-r)^{2}}\left(\frac{C_{r+1}}{C_{r}} \right)^{2}} = \sum_{r=0}^{n-1}{\frac{1}{(n-r)^{2}}\left(n-r\right)^{2}/(r+1)}
same as sandipan
= \sum_{r=0}^{n-1}{\frac{1}{r+1}}
now take summation of the series
that's wat i got ??
edited
11nCr+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@manmay you missed the square.... it will be summation 1/(1+r)2......
1conti....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@rvking i could'nt get what you are trying to tell....
how 11+122+132+....+1n2 = \pi2/6
please explain
1i 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.
62actually it is not \pi^{2/6} \text{ but } \pi^2/6 and as you all said this is a fact which we should not worry beyond a point at this time... just a fact.