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Solving equation of summation of n combinations


Exercise:

Solve the following equation nk=1(nk)(nk)=441

Solution: we have nk=1(nk)(nk)=441nk=1n(nk)nk=1k(nk)=441

We know that from Binomial theorem (x+y)n=nk=0(nk)xnkyk where kn

Put x=1&y=1 to get nk=0(nk)=2n(n0)+nk=1=2n where (n0)=1

So nk=1(nk)=2n1 thus nk=1n(nk)=nnk=1(nk)=n(2n1)=n2nn....()

Notice that, (nk)=n!(nk)!k!=n(n1)!k(k1)!(((n1)(k1))!=nk(n1k1)

hence k(nk)=n(n1k1)nk=1k(nk)=n1k=0n(n1k1)=n2n1....()

Thus nk=1n(nk)nk=1k(nk)=n2nnn2n1=441

n2nnn2n2=4412n(nn2)=441+n2n(n2)=441+n

n2n=882+2nn(2n2)=882

By using numerical method we get n=7




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