Question

evaluate the following sum.
\\(\displaystyle \sum_{k=1}^{3} \frac{k^3}{9} + \left( \displaystyle \sum_{k=1}^{3} k \
ight)^2\\)
\\(\displaystyle \sum_{k=1}^{3} \frac{k^3}{9} + \left( \displaystyle \sum_{k=1}^{3} k \
ight)^2 = \square\\)
(simplify your answer.)

Explanation:

Step1: Evaluate the first sum $\sum_{k = 1}^{3}\frac{k^{3}}{9}$

For $k = 1$: $\frac{1^{3}}{9}=\frac{1}{9}$
For $k = 2$: $\frac{2^{3}}{9}=\frac{8}{9}$
For $k = 3$: $\frac{3^{3}}{9}=\frac{27}{9} = 3$
Sum: $\frac{1}{9}+\frac{8}{9}+3=\frac{1 + 8}{9}+3=1 + 3=4$

Step2: Evaluate the second sum $\sum_{k = 1}^{3}k$ and square it

$\sum_{k = 1}^{3}k=1 + 2+3 = 6$
Square: $6^{2}=36$

Step3: Add the two results

$4+36 = 40$

Answer:

40