Question

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

Explanation:

Step1: Split the summation

We know that $\sum_{k = 1}^{n}(a_{k}+b_{k})=\sum_{k = 1}^{n}a_{k}+\sum_{k = 1}^{n}b_{k}$. So, $\sum_{k = 1}^{50}(1 + k^{2})=\sum_{k = 1}^{50}1+\sum_{k = 1}^{50}k^{2}$.

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

The sum of a constant $c$ from $k = 1$ to $n$ is $c\times n$. Here, $c = 1$ and $n=50$, so $\sum_{k = 1}^{50}1=1\times50 = 50$.

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 $n$ into the formula:

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Step4: Combine the two sums

Now, we add the two results together: $\sum_{k = 1}^{50}(1 + k^{2})=\sum_{k = 1}^{50}1+\sum_{k = 1}^{50}k^{2}=50+42925=42975$.

Answer:

42975