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Question

Find the sum of :
3.nC08.nC1+13.nC218.nC3....upto(n+1)terms

A
zero
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B
n+1
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C
n1
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D
(n+1)2
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Solution

The correct option is A zero
(1x)n=C0C1x+C2x2+....+(1)nCnxn ....(1)
(1x5)n=C0C1x5+C2x10+....+(1)nCnx5n
Differentiating wrt x
5nx4(1x5)n1=5C1x4+10C2x9+15C2x14+....+(1)nCnx5n
Multiplying (1) by 3 and then adding with (2):
3(1x)n5nx4(1x5)n1=3.nC08nC1x4+13nC2x918nC3x14....
now, put x=0
0=3nC08nC1+13nC218nC3....

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