Binomial Theorem 1 Question 17

18. The sum of coefficients of integral powers of x in the binomial expansion (12x)50 is

(2015 Main)

(a) 12(350+1)

(b) 12(350)

(c) 12(3501)

(d) 12(250+1)

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Answer:

Correct Answer: 18. (8)

Solution:

  1. Let Tt+1 be the general term in the expension of (12x)50

Tr+1=50Cr(1)50r(2x1/2)r=50Cr2rxr/2(1)r

For the integral power of x,r should be even integer.

Sum of coefficients =r=02550C2r(2)2r

=12[(1+2)50+(12)50]=12(350+1)

Alternate Solution

We have,

(12x)50=CoC12x+C2(2x)2++C50(2x)50(1+2x)50=Co+C12x+C2(2x)2++C50(2x)50

On adding Eqs. (i) and (ii), we get

(12x)50+(1+2x)50=2[C0+C2(2x)2++C50(2x)50](12x)50+(1+2x)502=C0+C2(2x)2++C50(2x)50

On putting x=1, we get

(121)50+(1+21)502=C0+C2++C50(2)50(1)50+(3)502=C0+C2(2)2++C50(2)501+3502=C0+C2(2)2++C50(2)50



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