Binomial Theorem 1 Question 18

19. Coefficient of x11 in the expansion of (1+x2)4(1+x3)7(1+x4)12 is

(2014 Adv.)

(a) 1051

(b) 1106

(c) 1113

(d) 1120

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Answer:

Correct Answer: 19. (n=6)

Solution:

  1. Coefficient of xr in (1+x)n is nCr.

In this type of questions, we find different composition of terms where product will give us x11.

Now, consider the following cases for x11 in

(1+x2)4(1+x3)7(1+x4)12.

Coefficient of x0x3x8; Coefficient of x2x9x0

Coefficient of x4x3x4; Coefficient of x8x3x0

=4C0×7C1×12C2+4C1×7C3×12C0+4C2×7C1

=462+140+504+7=1113

×12C1+4C4×7C1×12C0

x+1x2/3x1/3+1(x1)xx1/210

=(x1/3+1)(x+1)x10=(x1/3x1/2)10

The general term is

Tr+1=10Cr(x1/3)10r(x1/2)r=10Cr(1)rx10r3r2

For independent of x, put

10r3r2=0202r3r=020=5rr=4T5=10C4=10×9×8×74×3×2×1=210



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