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Question

If in the expansion of (1+x)n,the coefficient of pth and qth terms are equal and the relation between p and q is p-q =n . Find the value of p2+q2. where pq.

Or

Find the term independent of x in the expansion of (2x1x)10.


Solution

We know that the cofficient of rth term in the expansion of (1+x)n is nCr1.

Therefore, coefficients of pth and qth terms in the expansion of (1+x)n are nCp1., respectively.

It is given that these coefficients are equal.

nCp1.=nCq1

pa+q1=n            [nCx=nCyx+y=n]

P+q=n+2                  (i)

Also given,pq=n              (ii)

On adding Eqs,(i) and (ii) we get         

2p=2n+2p=n+1

On putting p=n+1 in Eq.(i), we get

                 q=n+2(n+1)=1

Now, p2+q2=(N+1)2+12

=n2+1+2n+1=n2+2n+2

Or

Let (r+1) th term be independent of x in the given expansion (2x1x)10.

Then , Tr+1=10Cr(2x)10r(1x)r

 Tr+1=(1)r 10Cr(2)10r(X)102r

For this term to be independent of x, we must have

102r=0r=5

So, (5+1) i.e. 6 th terms is independent of x, i.e T6=(1)5 10C5(2)105

=10!5!5!×25=252×32=8064


Mathematics

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