Question

Show that the (Summation from y=k-1 to 2k ) ( y) = (k+2)(3k-1)/2

Answer 1

K greater equal than 1

Answer 2

\[\sum_{y=k-1}^{2k}y=\frac{(k+2)(3k-1)}{2}\\
\text{Let }S=\sum_{y=k-1}^{2k}y\]
\[\begin{align*}&S=(k-1)+k+(k+1)+\cdots+(2k-2)+(2k-1)+2k\\
+&S=2k+(2k-1)+(2k-2)+\cdots+(k+1)+k+(k-1)\\
&--------------------------\\
&2S=(2k+k-1)+(2k+k-1)+\cdots+(2k+k-1)\end{align*}\]
On the RHS you have (2k - (k-2)) = (k+2) terms. Do you see why?
This gives you
\[2S=(k+2)(2k+k-1)\\
2S=(k+2)(3k-1)\\
S=\frac{(k+2)(3k-1)}{2}\\
\text{Thus, }\sum_{y=k-1}^{2k}y=\frac{(k+2)(3k-1)}{2}\]