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`f(n)=sum_(r=1)^(n) [r^(2)(""^(n)C_(r)-""^(n)C_(r-1))+(2r+1)(""^(n)C_(r ))]`, thenA. `f(30)=960`B. `f(21)=483`C. `f(16)=64`D. `f(11)=44` |
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Answer» Correct Answer - A::B `f(n)=underset(r=1)overset(n)sum[r^(2)(.^nC_(r)-.^nC_(r=1))+(2r).^nC_(r)+.^nC_(r)]` `=underset(r-1)overset(n)sum[(r^(2)+2r+1).^nC_(r)-r^(2).^nC_(r-1)]` `=underset(r=1)overset(n)sum[(r+1)^(2).^nC_(r)-r^(2).^nC_(r=1)]=underset(r=1)overset(n)sum[V_(r-1)-V_(r)]` `=V_(2)-V_(1)+V_(3)-V_(2)+...+V_(n+1)-V_(n)` `=V_(n+1)-V_(1)=(n+1)^(2).^(n)C_(n)-1=n^(2)+2n.` |
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