Question

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

$\underset{k=1}{\overset{26}{\sum (3}}k-7)$

Short answer

The sum of this sequence is 871.

Step-by-step solution

Step 1. Write the given information. 

The sum of sequence:

$\underset{k=1}{\overset{26}{\sum (3k-7)}}$

Step 2. Use the properties of sequence.

Using the properties of sequence:

$$\begin{gathered} \underset{k=1}{\overset{n}{\sum (a_k-b_k)}}=\underset{k=1}{\overset{n}{\sum \left(a_k\right)}}-\underset{k=1}{\overset{n}{\sum \left(b_k\right)}} \\ \& \\ \underset{k=1}{\overset{n}{\sum \left(c{a}_k\right)}}=c\underset{k=1}{\overset{n}{\sum a_k}} \end{gathered}$$

So,

$\underset{k=1}{\overset{26}{\sum (3k-7)}}=\underset{k=1}{\overset{26}{\sum \left(3k\right)}}-\underset{k=1}{\overset{26}{\sum \left(7\right)}}$

&

$\underset{k=1}{\overset{26}{\sum \left(3k\right)}}=\underset{k=1}{\overset{26}{3\sum \left(k\right)}}$

Step 3. Use the formula for sum of sequences of n real numbers.

Using formula for summation is:

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

So,

role="math" localid="1646750162806" $$\begin{gathered} 3\underset{k=1}{\overset{26}{\sum k}}=3\times \frac{26(26+1)}{2} \\ \Rightarrow \underset{k=1}{\overset{26}{3\sum k}}=3\times \frac{26\left(27\right)}{2} \\ \Rightarrow \underset{k=1}{\overset{26}{3\sum k}}=3\times 351 \\ \Rightarrow \underset{k=1}{\overset{26}{3\sum k}}=1053 \end{gathered}$$

Step 4. Use the formula for sum of sequence. 

Using formula for summation is:
$$\begin{gathered} \underset{k=1}{\overset{n}{\sum c}}=c+c+...+c \\ \Rightarrow \underset{k=1}{\overset{n}{\sum c}}=cn,c-realnumber \\ So, \\ \underset{k=1}{\overset{26}{\sum 7}}=7\times 26 \\ \Rightarrow \underset{k=1}{\overset{26}{\sum 7}}=182 \end{gathered}$$

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

From Step 3,

$\underset{k=1}{\overset{26}{3\sum k}}=1053$

From Step 4,

$\underset{k=1}{\overset{26}{\sum 7}}=182$

So,

$$\begin{gathered} \underset{k=1}{\overset{26}{\sum (3k-7)}}=\underset{k=1}{\overset{26}{\sum \left(3k\right)}}\underset{k=1}{\overset{26}{-\sum \left(3\right)}} \\ \Rightarrow \underset{k=1}{\overset{26}{\sum (3k-7)}}=1053-182 \\ \Rightarrow \underset{k=1}{\overset{26}{\sum (3k-7)}}=871 \end{gathered}$$