13n x (nCn-1) ....?
5 Answers
13n(1+x)n-1 = nC1 + 2 nC2 x +.........+ n.nCnxn-1
(1+x)n= nC0xn + nC1xn-1 + nC2xn-2+....+ nCnx0
Required Series = Multiplying the 2 Eq.n(s) & fnding the Coefficient of x n-1 = Coeff of x n-1 in n.(1+x)n = n.nCn-1
156no aveek..
far from it ;)
Think of the derivative of (1+x)n*(1+x)n
I am seeing a good innovative problem on this topic after a long long time :)
11Since (1+x) n = nC0 + n C1 x +n C 2 x2 + n C 3 x 3+.........+ n C n xn
Differentiating both sides w.r.t x, we get
n (1+x) n-1 = 0 +n C 1 +2 n C 2 x +3 n C 3 x2+.....+n nCn x n-1 ............................(i)
and (x+1) n = n C0 x n +nC1 x n-1+ n C2 x n-2 +n C3 xn-3+................+nCn ......................(ii)
Multiplying (i) and (ii), we get
n(1+x) 2n-1 = (n C 1 +2 n C 2 x +3 n C 3 x2+.....+n nCn x n-1 ) X ( n C0 x n +nC1 x n-1+ n C2 x n-2 +n C3 xn-3+................+nCn )
...................(iii)
Now coeff of x n-1 on R.H.S = (n C1) 2 +2( n C2 )2 +3 (nC3) 2 +.................n (nCn) 2
and coeff of x n-1 on L.H.S = n 2n-1 C n-1
= (2n-1) !{(n-1)!} 2
but (iii) is an identity, therefore, coeff of x n-1 in R.H.S = coeff of x n-1 in L.H.S
Therefore,
(n C1) 2 +2( n C2 )2 +3 (nC3) 2 +.................n (nCn) 2 = (2n-1) !{(n-1)!} 2
