If f:N Z is defined by f(n)=vmatrix n -1 -5 -2n^2 3(2k+1) 2k+1 -3n^3 3k(2k+1) 3k(k+2)+1 vm…

Question

If $f:N	o Z$ is defined by

$f(n)=\begin{vmatrix} n \quad  -1 \quad  -5 \ -2{n}^{2} \quad  3(2k+1) \quad  2k+1 \ -3{n}^{3} \quad  3k(2k+1) \quad  3k(k+2)+1 \end{vmatrix},k\in N,$

and $\sum _{n=1}^{k}f(n)=98$, then $k$ is equal to :

Accepted Answer

$f(n)=\begin{vmatrix} n \quad  -1 \quad  -5 \ -2{n}^{2} \quad  3(2k+1) \quad  2k+1 \ -3{n}^{3} \quad  3k(2k+1) \quad  3k(k+2)+1 \end{vmatrix}$

$\sum _{n=1}^{k}f(n)=\begin{vmatrix} \frac{k(k+1)}{2} \quad  -1 \quad  -5 \ -2\frac{k(k+1)(2k+1)}{6} \quad  3(2k+1) \quad  2k+1 \ -3\frac{{k}^{2}(k+1{)}^{2}}{4} \quad  3k(2k+1) \quad  3k(k+2)+1 \end{vmatrix}=98$

$\begin{gathered} \Rightarrow \frac{k(k+1)(2k+1)}{2}\cdot \frac{7}{3}=98 \ \Rightarrow k(k+1)(2k+1)=84 \ \Rightarrow k=3 \end{gathered}$

Solutions · 1