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Q 76.

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

The sum of this sequence is 871.

See the step by step solution

Step 1. Write the given information.

The sum of sequence:

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

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,

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

&

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

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

Using formula for summation is:

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

So,

role="math" localid="1646750162806" $3\underset{k=1}{\overset{26}{\sum k}}=3×\frac{26\left(26+1\right)}{2}\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{26}{3\sum k}}=3×\frac{26\left(27\right)}{2}\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{26}{3\sum k}}=3×351\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{26}{3\sum k}}=1053$

Step 4. Use the formula for sum of sequence.

Using formula for summation is:$\underset{k=1}{\overset{n}{\sum c}}=c+c+...+c\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{n}{\sum c}}=cn,c-realnumber\phantom{\rule{0ex}{0ex}}So,\phantom{\rule{0ex}{0ex}}\underset{k=1}{\overset{26}{\sum 7}}=7×26\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{26}{\sum 7}}=182\phantom{\rule{0ex}{0ex}}$

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,

$\underset{k=1}{\overset{26}{\sum \left(3k-7\right)}}=\underset{k=1}{\overset{26}{\sum \left(3k\right)}}\underset{k=1}{\overset{26}{-\sum \left(3\right)}}\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{26}{\sum \left(3k-7\right)}}=1053-182\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{26}{\sum \left(3k-7\right)}}=871$