Question

# The coefficient of x in the expansion (1+x)(1+2x)(1+3x)⋯(1+100x) is also equal to

A
1002992+982972+962952++2212
B
The sum of all 101 A.M.'s inserted between 1 and 99
C
The sum of all 100 A.M.'s inserted between 1 and 99
D
The sum of first 20 terms of the series 1+(1+3)+(1+3+5)+(1+3+5+7)+

Solution

## The correct options are A 1002−992+982−972+962−952+⋯+22−12 B The sum of all 101 A.M.'s inserted between 1 and 99 Coefficient of x in the expansion (1+x)(1+2x)(1+3x)⋯(1+100x) =1+2+3+⋯+100=100×1012=5050 Now, 1002−992+982−972+962−952+…+22−12=(1002−992)+(982−972)+(962−952)+…+(22−12)=(100+99)(100−99)+(98+97)(98−97)+…+(2+1)(2−1)=100+99+98+97+…+2+1=5050 Now, Let n A.M.'s are inserted between 1 and 99, then the sum of them  5050=n+22[1+99]−(1+99)⇒5050=100[n+22−1]⇒5050=50n⇒n=101 Therefore 101 A.M.'s are inserted. Now, the sum of first 20 terms of the series 1+(1+3)+(1+3+5)+(1+3+5+7)+⋯  General term of the series is Tr=1+3+5+7⋯(2r−1)=r2⇒S=20∑r=1Tr=20∑r=1r2⇒S=n(n+1)(2n+1)6⇒S=20×21×416=2870Mathematics

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