Question

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

$\underset{k=0}{\overset{14}{\sum (k^2}}-4)$

Short answer

The sum of this sequence is 955.

Step-by-step solution

Step 1. Write the given information. 

The sum of sequence:

$\underset{k=0}{\overset{14}{\sum (k^2-4)}}$

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,

$\underset{k=1}{\overset{14}{\sum (k^2-4)}}=\underset{k=1}{\overset{14}{\sum \left(k^2\right)}}\underset{k=1}{\overset{14}{-\sum \left(4\right)}}$

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

Using formula for summation is:

$$\begin{gathered} \underset{k=0}{\overset{n}{\sum k^2}}=1^2+2^2+...+n^2 \\ \underset{k=0}{\overset{n}{\sum k^2}}=0+\underset{k=1}{\overset{n}{\sum k^2}} \\ \Rightarrow \underset{k=0}{\overset{n}{\sum k^2}}=0+\frac{n(n+1)(2n+1)}{6} \end{gathered}$$

So,

$$\begin{gathered} \underset{k=0}{\overset{14}{\sum k^2}}=\frac{14(14+1)\left(\right(2\times 14)+1)}{6} \\ \Rightarrow \underset{k=0}{\overset{14}{\sum k^2}}=\frac{6090}{6} \\ \Rightarrow \underset{k=0}{\overset{14}{\sum k^2}}=1015 \end{gathered}$$

Step 4. Use the formula for sum of sequences. 

Using formula for summation is:

$$\begin{gathered} \underset{k=0}{\overset{n}{\sum c}}=c+c+...+c \\ \Rightarrow \underset{k=0}{\overset{n}{\sum c}}=cn,c-realnumber \\ So, \\ \underset{k=0}{\overset{14}{\sum 4}}=4\times (14+1) \\ \Rightarrow \underset{k=0}{\overset{14}{\sum 4}}=4\times 15 \\ \Rightarrow \underset{k=0}{\overset{14}{\sum 4}}=60 \end{gathered}$$

Step 5. Now, subtract Step 3  from Step 4.

From Step 3,

$\underset{k=1}{\overset{14}{\sum k^2}}=1015$

From Step 4,

$\underset{k=1}{\overset{14}{\sum 4}}=60$

So,

$$\begin{gathered} \underset{k=1}{\overset{14}{\sum (k^2-4)}}=\underset{k=1}{\overset{14}{\sum \left(k^2\right)}}\underset{k=1}{\overset{14}{-\sum \left(4\right)}} \\ \Rightarrow \underset{k=1}{\overset{14}{\sum (k^2-4)}}=1015-60 \\ \Rightarrow \underset{k=1}{\overset{14}{\sum (k^2-4)}}=955 \end{gathered}$$