Binomial Theorem

q1 S=r=0n1(nr)21(CrCr+1)2

q2 S=n(n+2)C0(n+1)(n+3)C1+(n+2)(n+4)C2......(n+1)terms

9 Answers

1
Manmay kumar Mohanty ·

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
01xn1(1x)ndx=nC0n+1C1+n+2C2............

to integrate L.H.S substitute x = sin2θ Integral reduces to 20π/2sin2n2θcos2n+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=(n1)! if nϵN hence integral reduces to
20π/2sin2n2θcos2n+1θdθ = 2Γ(22n1+2n+1+2)2.Γ(22n1+1)Γ(22n+1+1)=Γ(2n+1)ΓnΓ(n+1)=2n!(n1)!n!

wat to do next can't thnk
[7][7]

11
SANDIPAN CHAKRABORTY ·

for Q1 i'm half done..

r=0n1(nr)21(CrCr+1)2

r=0n1(nr)21(1+rn(1+r)+1)2

r=0n1(nr)21(1+rnr)2

r=0n1(r+1)21

after this i got stuck [12][12][12][12]

1
Manmay kumar Mohanty ·

Q1) CrCr+1=(nr)!r!n!(n(r1))!(r+1)!n!=(n(r1))!(nr)!=(nr)

so r=0n1(nr)21(CrCr+1)2 = r=0n1(nr)21(nr)2/(r+1)
same as sandipan
= r=0n1r+11
now take summation of the series

that's wat i got ??

edited

11
SANDIPAN CHAKRABORTY ·

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 ·

@manmay you missed the square.... it will be summation 1/(1+r)2......

1
rvking ·

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 ·

@rvking i could'nt get what you are trying to tell....

how 11+122+132+....+1n2 = π2/6

please explain

1
rvking ·

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 ·

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.

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