If ^20C_1+(2^2) ^20C_2+(3^2) ^20C_3+.....+20^2 ^20C_20=A2^(β), then the ordered pair (A,β) is equal to

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MARKS Content Team · Accepted Answer

Correct answer: (B) (420, 18). (1+x)^n= ^nC_0+ ^nC_1x+ ^nC_2x^2+......+ ^nC_nx^n Diff. w. r. t. x ⇒n(1+x)^(n-1)= ^nC_1 + ^nC_2(2x)+.....+ ^nC_nn(x)^(n-1) Multiply by x both side ⇒nx(1+x)^(n-1)= ^nC_1x+ ^nC_22x^2+.....+ ^nC_nnx^n Diff. w. r. t. x ⇒n[(1+x)^(n-1)+(n-1)x(1+x)^(n-2)] = ^nC_1 + ^nC_22^2X+......+ ^nC_n(n^2)x^(n-1) Put x=1 and n=20 ⇒ ^20C_1+ 2^2 ^20C_2+ 3^2 ^20C_3+......+(20)^2 ^20C_20 =20×2^182+19=4202^18=A2^(β) Hence, A,β=(420, 18)

If $^{20} C_{1} + (2^{2})^{20} C_{2} + (3^{2})^{20} C_{3} + . . . . . + (20^{2})^{20} C_{20} = A (2^{β}),$ then the ordered pair $(A, β)$ is equal to

Solution

Relevant Questions

If 20C1+(22) 20C2+(32) 20C3+.....+202 20C20=A2β, then the ordered pair (A,β) is equal to

Solution

Relevant Questions

If $^{20} C_{1} + (2^{2})^{20} C_{2} + (3^{2})^{20} C_{3} + . . . . . + (20^{2})^{20} C_{20} = A (2^{β}),$ then the ordered pair $(A, β)$ is equal to