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

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

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: