Find sum of (1^2)*28C1+(2^2)*28C2+.........upto(28^2)28C28.

• ans is 406.2^27
  We know,(1+x)^n=C0+C1x+C2x^2+.....+Cnx^n .......(1)
  Differentiating w.r.t x we get
  n(1+x)^(n−1)=C1+2C2x+3C3x^2+.....+nCnx^(n−1)
  Multiplying by x on both sides,
  x.n(1+x)^(n−1)=x.C1+2C2x^2+3C3x^3+.....
  Now again differentiating w.r.t to x
  n(1+x)^(n−1)+n(n−1)x(1+x)^(n−2)=C0+2^2C1x+3^2C2x^2+4^2C3x^3+.....
  Putting x = 1, we get
  n(n+1)2^(n−2)=C1+2^2C2+3^2C3+4^2C4+.....
  Now substituting n=28
  28(28+1)2^(28−2) = 812.2^26 = 406.2^27
• 11 years agoHelpfull: Yes(31) No(1)
• sum=n/2(a+l)
  a:first term
  l:last term
  sum=11368
• 11 years agoHelpfull: Yes(0) No(3)