Question

The value of  \[\sum^n_{r = 1} \left\{ (2r - 1) a + \frac{1}{b^r} \right\}\] is equal to

Options

• \[a n^2 + \frac{b^{n - 1} - 1}{b^{n - 1} (b - 1)}\]

• \[a n^2 + \frac{b^n - 1}{b^n (b - 1)}\]

• \[a n^3 + \frac{b^{n - 1} - 1}{b^n (b - 1)}\]

• none of these

Answer 1

\[a n^2 + \frac{b^n - 1}{b^n (b - 1)}\]

We have:

\[\sum^n_{r = 1} \left\{ (2r - 1) a + \frac{1}{b^r} \right\}\]

\[ = \sum^n_{r = 1} \left\{ 2ra - a + \frac{1}{b^r} \right\}\]

\[ = \sum^n_{r = 1} 2ar - \sum^n_{r = 1} a + \sum^n_{r = 1} \frac{1}{b^r}\]

\[ = an\left( n + 1 \right) - an + \frac{\left( 1 - b^n \right)}{\left( 1 - b \right) b^n}\]

\[ = a n^2 + \frac{\left( b^n - 1 \right)}{\left( b - 1 \right) b^n}\]