Q.

 nC04nCnnC1(4n3)Cn+nC24n6Cn_nC34n9Cn+...+  to (n + 1) terms is equal to (1+k)n;  then the value of k is  (kN)

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answer is 2.

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Detailed Solution

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Coefficient xn in   {nC0(1+x)4n_nC1(1+x)4n3+nC2(1+x)4n6......+(1)nnCn(1+x)n}
 Coefficient  xn in  (1+x)n{(1+x)31}n3n

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 nC0 4nCn−nC1 (4n−3)Cn+nC2 4n−6Cn_nC3 4n−9Cn+...+  to (n + 1) terms is equal to (1+k)n;  then the value of k is  (k∈N)