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Question

Prove that the coefficient of (r+1)th term in the expansion of (1+x)n+1 is equal to the sum of the coefficients of rth and (r+1)th terms in the expansion of (1+x)n.


Solution

Now, coefficient of (r+1)th term in the expansion of (4+x)n+1=n+1Cr+11

=n+1Cr and, coefficient of rth term in (1+x)n +coefficient of (r+1)th term in (1+x)n

=nCr1+nCr+11

=nCr1+nCr

=n!{n(r1)} (r1)!+n!(nr)!r!

=n!{nr+1}!(r1)!+n!(nr)!r!

=n!(nr+1)(nr)!(r1)!+n!(nr)!r(r1)!

=n!(nr+1)(nr)!(r1)!+n!(nr)!(r1)!r

=n!(nr)!(r1)![1nr+1+1r]

=n!(nr)!(r1)![r+nr+1(nr+1)r]

=n!(nr)!(r1)![n+1(nr+1)r]

=n!(n+1)(nr)!(nr+1)(r1)!r

=(n+1)!(nr+1)!r!

=(n+1)!(n+1r)!r!

=n+1Cr

n+1Cr=nCr1+nCr

The coefficient of (r+1)th term in the expansion of (1+x)n+1 is equal to the sum of the coefficient of rth and (r+1)th terms in the expansion of (1+x)n.


Mathematics
RD Sharma
Standard XI

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