Find each sum:

(a) $\displaystyle\sum\limits_{k=1}^{100}\dfrac{1}{2}$

(b) $\displaystyle\sum\limits_{k=1}^{5} ( 3k)$

(c) $\displaystyle\sum\limits_{k=1}^{10}( k^{3}+1)$

(d) $\displaystyle\sum\limits_{k=6}^{20} ( 4k^{2})$

Solution (a) $\displaystyle\sum\limits_{k=1}^{100}\dfrac{1}{2}\underset{\underset{\color{#0066A7}{\hbox{(1)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} \frac{1}{2}\textstyle\displaystyle\sum\limits_{k=1}^{100}1\underset{\underset{\color{#0066A7}{\hbox{(5)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} \dfrac{1}{2}\cdot 100=50$

A-42

(b) $\displaystyle\sum\limits_{k=1}^{5} ( 3k) \underset{\underset{\color{#0066A7}{\hbox{(1)}}}{\color{#0066A7}{\displaystyle\uparrow}}}{=} 3\displaystyle\sum\limits_{k=1}^{5}k\underset{\underset{\color{#0066A7}{\hbox{(6)}}}{\color{#0066A7}{\displaystyle\uparrow}}}{=} 3\left( \dfrac{5 ( 5+1) }{2}\right)=3 ( 15) =45$

(c)$\displaystyle\sum\limits_{k=1}^{10} ( k^{3}+1) \underset{\underset{\color{#0066A7}{\hbox{(2)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} \displaystyle\sum \limits_{k=1}^{10}k^{3}+\displaystyle\sum\limits_{k=1}^{10}1 \underset{\underset{\color{#0066A7}{\hbox{(8) and (5)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} \left[ \dfrac{10( 10+1) }{2}\right] ^{2} +10 =3025+10=3035$

(d)Notice that the index of summation starts at 6. $$\begin{eqnarray*} \sum\limits_{k=6}^{20} ( 4k^{2}) &\underset{\underset{\color{#0066A7}{\hbox{(1)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=}& 4\sum\limits_{k=6}^{20}k^{2}\underset{\underset{\color{#0066A7}{\hbox{(4)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} 4\left[ \sum\limits_{k=1}^{20}k^{2}-\sum \limits_{k=1}^{5}k^{2}\right] \nonumber \\ & \underset{\underset{\color{#0066A7}{\hbox{(7)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} & 4\left[ \dfrac{20(21) ( 41) }{6}-\dfrac{5( 6) ( 11) }{6}\right] =4[ 2870-55] =11,260 \\[-10.5pt] \end{eqnarray*}$$