Evaluate \displaystyle\sum_{k=1}^{11} (2+3k)

\( \displaystyle\sum_{k=1}^{11} (2+3k)=\displaystyle\sum_{k=1}^{11} (2)+\)

\( \displaystyle\sum_{k=1}^{11} (3k)=22\displaystyle\sum_{k=1}^{11} (3^k)\) .......(1)

We know that,

\( \displaystyle\sum_{k=1}^{11} (3^k)=3^1+3^2+....+3^{11}\)

This sequence \( 3,,3^2,3^3....3^{11}\) forms a G.P. Therefore,

\( s_n=\dfrac{a(1-r^n)}{1-r}\)

Substituting the values to the above equation we get,

\( s_n=\dfrac{3[(3)^{11}-1]}{3-1}\)

\( s_n=\dfrac{3}{2}(3^{11}-1)\)

Therefore,

\( \displaystyle\sum_{k=1}^{11} (3^k)=\dfrac{3}{2}(3^{11}-1)\)

Substitute this value in equation (1).

\( \displaystyle\sum_{k=1}^{11} (2+3k)=22+\dfrac{3}{2}(3^{11}-1)\)