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If ( 1+ x) n = 1+ n C 1 x + n C 2 x 2 + ... + n C n x n and a 0 , a 1 , a 2 , a 3 , ..., a n are in A. P., show that a 0 - n C 1 .a 1 + n C 2 .a 2 ... + (-1) n . n C n .a n = 0.

If ( 1+ x)n = 1+ nC1x + nC2x2 + ... + nCnxn and a0, a1, a2, a3, ..., an are in A. P., show that a- nC1.a1 + nC2.a2... + (-1)n.nCn.an = 0.

Grade:11

1 Answers

Aditya Gupta
2081 Points
5 years ago
General term is
(-1)^r*(a0+(r-1)d)*nCr
= (-1)^r*(a0-d)*nCr + (-1)^r*dr*nCr
Now in the first sum, a0-d is a constant so it comes out and we are left with (-1)^r*nCr whose sum is zero.
For the second sum, we have (-1)^r*r*nCr. But rnCr= n*(n-1)C(r-1) so sum becomes (-1)^rn-1Cr-1 which is also zero.

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