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Question

For n>0, the value of nC022 nC1+32 nC242 nC3 upto (n+1) terms is kn(n+1)2n, then k=

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Solution

Let S= nC022 nC1+32 nC242 nC3+ upto (n+1) terms
S=nr=0(1)r(r+1)2 nCr
=nr=0(1)r(r2+2r+1) nCr
=nr=0(1)r(r(r1)+3r+1) nCr
=nr=0(1)rr(r1) nCr+3nr=0(1)rrCr+nr=0(1)r nCr
=nr=2(1)rr(r1) nCr+3nr=1(1)rrCr+nr=0(1)r nCr
=n(n1)nr=2(1)r n2Cr2+3nnr=1(1)r n1Cr1+(11)n
=n(n1)nr=2 n2Cr2(1)r23nnr=1 n1Cr1(1)r1+0
=0 [nr=0ncrar=(1+a)n]

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