 QUESTION

# Find out the sum of the series up to n terms.$\left( {a + b} \right) + \left( {{a^2} + 2b} \right) + \left( {{a^3} + 3b} \right) + ...........$ to n terms.

Hint: Determine the type of series and use the corresponding summation formula.

${{\text{n}}^{th}}$ term of the series is $\left( {{a^n} + nb} \right) \\$

Series is written as;

$\Rightarrow \left( {a + b} \right) + \left( {{a^2} + 2b} \right) + \left( {{a^3} + 3b} \right) + ........... + \left( {{a^n} + nb} \right) \\$

Now rearrange the series,

$\Rightarrow S = \left( {a + {a^2} + {a^3} + {a^4}.............{a^n}} \right) + b\left( {1 + 2 + 3 + 4 + ...............n} \right) \\$

Now it is clear that the first series is in G.P. with first term, ${{a_1}} = a$ and the common ratio, $r = \dfrac{{{a^2}}}{a} = a \\$ and last term is ${a}^n$.

Sum of n terms of G.P. is $\left( {{S_n}} \right) = {a_1}\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right) = a\left( {\dfrac{{{a^n} - 1}}{{a - 1}}} \right) \\$

The next series summation is $b\sum\limits_{k = 1}^n k = b\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right) \\$

$\Rightarrow S = a\left( {\dfrac{{{a^n} - 1}}{{a - 1}}} \right) + b\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right) \\$

So, this is our required answer.

Note: Such problems can be easily solved by identifying the series and using the corresponding summation formula. The second part of the series can also be solved by considering it as an A.P. and then finding the sum.