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Suppose det ` [{:(sum_(k=0)^(n)k,,sum_(k=0)^(n).^nC_(k)k^2),(sum_(k=0)^(n).^nC_(k)k,,sum_(k=0)^(n).^nC_(k)3^(k)):}]=0` holds for some positive integer n. then `sum_(k=0)^(n)(.^nC_(k))/(k+1)` equals ............ |
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Answer» Correct Answer - `6.20` It is given that `| {:(Sigma_(k=0)^(n)k,Sigma_(k=0)^(n).^nC_kk^2),(Sigma_(k=0)^(n).^nC_k.k,Sigma_(k=0)^(n).^nC_k3^k):}|=0` `| {:((n(n+1))/(2),n(n+1)2^(n-2)),(n.2^(n-1),4^n):}|=0` `[ {:(becauseSigma_(k=0)^(n)k=(n(n+1))/(2)",",Sigma_(k=0)^(n).^nC_k k=n.2^(n-1)),(Sigma_(k=0)^(n).^nC_k k^2=n(n+1)2^(n-1)and,Sigma_(k=0)^(n).^nC_k 3^(k)=4^n):}]` `rArr (n(n+1))/(2)4^n-n^2(n+1)2^(2n-3)=0` ` rArr (4^n)/(2)-n(4^(n-1))/(2)=0rArr n=4` ` therefore Sigma_(k=0)^(n)(.^nC_(k))/(k+1)=Sigma_(k=0)^(4)(.^4C_k)/(k+1)=(1)/(5)Sigma_(k=0)^(4).^3C_k+1=(1)/(5)(2^5-1)=(1)/(5)(32-1)=(31)/(5)=6.20` |
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