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Q 75.

Expert-verified
Found in: Page 809

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# In Problems 71-82, find the sum of each sequence.$\underset{k=1}{\overset{20}{\sum \left(5}}k+3\right)$

The sum of this sequence is 1110.

See the step by step solution

## Step 1. Write the given information.

The sum of sequence:

$\underset{k=1}{\overset{20}{\sum \left(5k+3\right)}}$

## Step 2. Use the properties of sequence.

Using the properties of sequence: $\underset{k=1}{\overset{n}{\sum \left({a}_{k}+{b}_{k}\right)}}=\underset{k=1}{\overset{n}{\sum \left({a}_{k}\right)}}+\underset{k=1}{\overset{n}{\sum \left({b}_{k}\right)}}\phantom{\rule{0ex}{0ex}}&\phantom{\rule{0ex}{0ex}}\underset{k=1}{\overset{n}{\sum \left(c{a}_{k}\right)}}=c\underset{k=1}{\overset{n}{\sum {a}_{k}}}$

So,

$\underset{k=1}{\overset{20}{\sum \left(5k+3\right)}}=\underset{k=1}{\overset{20}{\sum \left(5k\right)}}+\underset{k=1}{\overset{20}{\sum \left(3\right)}}$

&

$\underset{k=1}{\overset{20}{\sum \left(5k\right)}}=\underset{k=1}{\overset{20}{5\sum \left(k\right)}}$

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

Using formula for summation is:

$\underset{k=1}{\overset{n}{\sum k}}=1+2+...+n\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{n}{\sum k}}=\frac{n\left(n+1\right)}{2}$

So,

$5\underset{k=1}{\overset{20}{\sum k}}=5×\frac{20\left(20+1\right)}{2}\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{n}{5\sum k}}=5×\frac{20\left(21\right)}{2}\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{n}{5\sum k}}=5×210\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{n}{5\sum k}}=1050$

## Step 4. Use the formula for sum of sequence.

Using formula for summation is:

$\underset{k=1}{\overset{n}{\sum c}}=c+c+...+c\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{n}{\sum c}}=cn,c-realnumber\phantom{\rule{0ex}{0ex}}So,\phantom{\rule{0ex}{0ex}}\underset{k=1}{\overset{20}{\sum 3}}=3×20\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{20}{\sum 3}}=60\phantom{\rule{0ex}{0ex}}$

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

From Step 3,

$\underset{k=1}{\overset{n}{5\sum k}}=1050$

From Step 4,

$\underset{k=1}{\overset{20}{\sum 3}}=60$

So,

localid="1646750229919" $\underset{k=1}{\overset{20}{\sum \left(5k+3\right)}}=\underset{k=1}{\overset{20}{\sum \left(5k\right)}}+\underset{k=1}{\overset{20}{\sum \left(3\right)}}\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{20}{\sum \left(5k+3\right)}}=1050+60\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{20}{\sum \left(5k+3\right)}}=1110$