evaluate the sum.
$\displaystyle\sum_{k=1}^{50} (1 + k^2)$
$\displaystyle\sum_{k=1}^{50} (1 + k^2) = 42975$ (type an integer or a simplified fraction.)

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Accepted Answer

# Explanation:
## Step1: Split the summation
We know that $\sum_{k = 1}^{n}(a + b_k)=\sum_{k = 1}^{n}a+\sum_{k = 1}^{n}b_k$. So for $\sum_{k = 1}^{50}(1 + k^2)$, we can split it into $\sum_{k = 1}^{50}1+\sum_{k = 1}^{50}k^2$.
## Step2: Evaluate $\sum_{k = 1}^{50}1$
The sum of a constant $c$ from $k = 1$ to $n$ is $n\times c$. Here $c = 1$ and $n = 50$, so $\sum_{k = 1}^{50}1=50\times1 = 50$.
## Step3: Evaluate $\sum_{k = 1}^{50}k^2$
The formula for the sum of squares of the first $n$ positive integers is $\sum_{k = 1}^{n}k^2=\frac{n(n + 1)(2n + 1)}{6}$. For $n = 50$, we substitute $n = 50$ into the formula:
$$
\begin{align*}
\sum_{k = 1}^{50}k^2&=\frac{50\times(50 + 1)\times(2\times50+ 1)}{6}\
&=\frac{50\times51\times101}{6}\
&=\frac{50\times51\times101}{6}\
&=\frac{2550\times101}{6}\
&=\frac{257550}{6}\
& = 42925
\end{align*}
$$
## Step4: Add the two sums
Now we add the results of the two sums: $\sum_{k = 1}^{50}1+\sum_{k = 1}^{50}k^2=50 + 42925=42975$.
# Answer:
$42975$

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QUESTION IMAGE

Question

Explanation:

Step1: Split the summation

Step2: Evaluate \(\sum_{k = 1}^{50}1\)

Step3: Evaluate \(\sum_{k = 1}^{50}k^2\)

Step4: Add the two sums

Answer: