The sum of coefficients of integral powers of r in the binomial expansion of (1−2x)50is

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12350+1

12350

12350−1

12250+1

answer is A.

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Detailed Solution

Let Tr+1 be the general term in the expansion of (1−2x)50∴ Tr+1=50Cr(1)50−r−2x1/2r=50Cr2rxn/2(−1)rFor the integral power of x, r should be even integer. Sum of coefficients =∑r=025 50C2r(2)2r =12(1+2)50+(1−2)50=12350+1Aliter We have,(1−2x)50=C0−C12x+C2(2x)2+…+C50(2x)50…(i) (1+2x)50=C0+C12x+C2(2x)2+…+C50(2x)50… (ii) On adding Eqs. (i) and (ii), we get(1−2x)50+(1+2x)50=2C0+C2(2x)2 +…+C50(2x)50⇒ (1−2x)50+(1+2x)502=C0+C2(2x)2+…+C50(2)50⇒ (−1)50+(3)502=C0+C2(2)2+…+C50(2)50⇒ 1+3502=C0+C2(2)2+…+C50(2)50

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