Question

Evaluate the sum of geometric series below:
$\displaystyle \large{ \sum_{k = 1}^n (4 \cdot {3}^{k} - {2}^{k + 1}) }$
Please also show your work too. Thank you!​

Answer 1

Answer:

$\displaystyle 6({3}^{n} - 1)- 4( {2}^{n} - 1 )$

Step-by-step explanation:

we would like to evaluate the following sum of geometric series:

$\displaystyle \sum_{k = 1}^n (4 \cdot {3}^{k} - {2}^{k + 1})$

to do so recall the Substracting property of partial sum Thus,

$\displaystyle \sum_{k = 1}^n (4 \cdot {3}^{k} )- \sum_{k = 1}^n({2}^{k + 1})$

rewrite $2^{k+1}$ as 2•2^k:

$\displaystyle \sum_{k = 1}^n (4 \cdot {3}^{k} )- \sum_{k = 1}^n({2}^{k } \cdot 2)$

utilize constant property of partial sum:

$\displaystyle 4\sum_{k = 1}^n ({3}^{k} )- 2\sum_{k = 1}^n({2}^{k } )$

factor out 2:

$\displaystyle 2 \left(2\sum_{k = 1}^n ({3}^{k} )- \sum_{k = 1}^n({2}^{k } ) \right)$

calculate the sum:

$\displaystyle 2 \left( \frac{ 6({3}^{n} - 1)}{2} - 2( {2}^{n} - 1 ) \right)$

distribute:

$\displaystyle \boxed{6({3}^{n} - 1)- 4( {2}^{n} - 1 ) }$

and we're done: