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

The correct options are
A 1002992+982972+962952++2212
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,
1002992+982972+962952++2212=(1002992)+(982972)+(962952)++(2212)=(100+99)(10099)+(98+97)(9897)++(2+1)(21)=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+221]5050=50nn=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(2r1)=r2S=20r=1Tr=20r=1r2S=n(n+1)(2n+1)6S=20×21×416=2870

Mathematics

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