Question

In Problems 71-82, find the sum of each sequence.

$\underset{k=1}{\overset{16}{\sum (k^2}}+4)$

Short answer

The sum of this sequence is 1560.

Step-by-step solution

Step 1. Write the given information. 

The sum of sequences:

$\underset{k=1}{\overset{16}{\sum (k^2+4)}}=\underset{k=1}{\overset{16}{\sum \left(k^2\right)}}+\underset{k=1}{\overset{16}{\sum \left(4\right)}}$

Step 2. Use the properties of sequence.

Using the properties of sequence:

$\underset{k=1}{\overset{n}{\sum (a_k+b_k)}}=\underset{k=1}{\overset{n}{\sum \left(a_k\right)}}+\underset{k=1}{\overset{n}{\sum \left(b_k\right)}}$

So,

$\underset{k=1}{\overset{16}{\sum (k^2+4)}}=\underset{k=1}{\overset{16}{\sum \left(k^2\right)}}+\underset{k=1}{\overset{16}{\sum \left(4\right)}}$

Step 3. Use the formula for sum of sequences of (n2 ) real numbers.

Using formula for summation is:

$$\begin{gathered} \underset{k=1}{\overset{n}{\sum k^2}}=1^2+2^2+...+n^2 \\ \Rightarrow \underset{k=1}{\overset{n}{\sum k^2}}=\frac{n(n+1)(2n+1)}{6} \end{gathered}$$

So,

$$\begin{gathered} \underset{k=1}{\overset{16}{\sum k^2}}=\frac{16(16+1)\left(\right(2\times 16)+1)}{6} \\ \Rightarrow \underset{k=1}{\overset{16}{\sum k^2}}=\frac{16\times 17\times 33}{6} \\ \Rightarrow \underset{k=1}{\overset{16}{\sum k^2}}=\frac{8976}{6} \\ \Rightarrow \underset{k=1}{\overset{16}{\sum k^2}}=1496 \end{gathered}$$

Step 4. Use the formula for sum of sequences. 

Using formula for summation is:
$$\begin{gathered} \underset{k=1}{\overset{n}{\sum c}}=c+c+...+c \\ \Rightarrow \underset{k=1}{\overset{n}{\sum c}}=cn,c-realnumber \\ So, \\ \underset{k=1}{\overset{16}{\sum 4}}=4\times 16 \\ \Rightarrow \underset{k=1}{\overset{16}{\sum 4}}=64 \end{gathered}$$

Step 5. Now, add the two sums from Step 3 and Step 4.

From Step 3,

$\underset{k=1}{\overset{16}{\sum k^2}}=1496$

From Step 4,

$\underset{k=1}{\overset{16}{\sum 4}}=64$

So,

$$\begin{gathered} \underset{k=1}{\overset{16}{\sum (k^2+4)}}=\underset{k=1}{\overset{16}{\sum \left(k^2\right)}}+\underset{k=1}{\overset{16}{\sum \left(4\right)}} \\ \Rightarrow \underset{k=1}{\overset{16}{\sum (k^2+4)}}=1496+64 \\ \Rightarrow \underset{k=1}{\overset{16}{\sum (k^2+4)}}=1560 \end{gathered}$$