If   20 C 1 + ( 2 2 )     20 C 2 + ( 3 2 )   20 C 3 + . . . . . + 20 2  …

If  20C1+(22)  20C2+(32) 20C3+.....+202  20C20=A2β, then the ordered pair (A,β) is equal to
  1. 380, 19
  2. (420, 18)
  3. (420, 19)
  4. (380, 18)

Solution

(1+x)n= nC0+ nC1x+ nC2x2+......+ nCnxn
Diff. w. r. t. x
⇒n(1+x)n-1= nC1 + nC2(2x)+.....+ nCnn(x)n-1
Multiply by x both side
⇒nx(1+x)n-1= nC1x+ nC22x2+.....+ nCnnxn
Diff. w. r. t. x
⇒n[(1+x)n-1+(n-1)x(1+x)n-2] = nC1 + nC222X+......+ nCn(n2)xn-1
Put x=1 and n=20
⇒ 20C1+ 22 20C2+  32 20C3+......+(20)2 20C20
=20×2182+19=420218=A2β
Hence, A,β=(420, 18)

Asked in: JEE Main 2019 (12 Apr Shift 2)

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