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Find: (nC0)^2+(nC1)^2+(nC2)^2+.......+(n...

Find: `(nC_0)^2+(nC_1)^2+(nC_2)^2+.......+(nC_n)^2`

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Prove that (2nC_0)^2-(2nC_1)^2+(2nC_2)^2+.....+(2nC_2n)^2=(-1)^n2nC_n

Prove that (2nC_0)^2+(2nC_1)^2+(2nC_2)^2-+(2nC_(2n))^2=(-1)^n2nC_ndot

If n=5 , then (^nC_(0))^(2)+(^nC_(1))^(2)+(^nC_(2))^(2)+....+(^nC_(5))^(2) is equal to

(nC_(0))^(2)-(nC_(1))^(2)+(nC_(2))^(2)+....+(-1)^(n)(nC_(n))^(2)

The expression nC_(0)+4*nC_(1)+4^(2)nC_(2)+.....4^(n^(n))C_(n), equals

The value of sum ( nC1)^2+( nC2)^2+( nC_3)^2+....+( nCn)^2 is

If (nC_0)/(2^n)+2.(nC_1)/2^n+3.(nC_2)/2^n+....(n+1)(nC_n)/2^n=16 then the value of 'n' is

If (nC_0)/(2^n)+2.(nC_1)/2^n+3.(nC_2)/2^n+....(n+1)(nC_n)/2^n=16 then the value of 'n' is

The value of ("^n C_0)/n + ("^nC_1)/(n+1) + ("^nC_2)/(n+2) +....+ ("^nC_ n)/(2n) is equal to

nC_r+2. nC_(r-1)+nC_(r-2)=