4.8 Q-2

Exercise 4.8 — Question 2

This question typically involves evaluating series using standard summation formulas or working with arithmetic-geometric sequences.

Key Formulas

Sum of first $n$ natural numbers: ${\sum }_{k=1}^{n}k=\frac{n(n+1)}{2}$

Sum of squares of first $n$ natural numbers: ${\sum }_{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$

Sum of cubes of first $n$ natural numbers: ${\sum }_{k=1}^n{k}^3={\left[\frac{n(n+1)}{2}\right]}^2$

Arithmetic-Geometric Series:

A series of the form $a+(a+d)r+(a+2d)r^2+\cdots$ is called an arithmetic-geometric series.

• General term: $T_n=[a+(n-1)d]r^{n-1}$
• Sum to $n$ terms is found using the subtraction method (multiply by $r$ and subtract).
• Sum to infinity (when $|r|<1$): $S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$

Worked Approach

To evaluate a series using sigma notation:

1. Identify whether the series is a pure summation (∑n, ∑n², ∑n³) or an arithmetic-geometric type.
2. Apply the appropriate closed-form formula.
3. Substitute the value of $n$ and simplify.

Example: Evaluate $\sum _{k=1}^n(2k^2+3k+1)$

$=2{\sum }_{k=1}^n{k}^2+3{\sum }_{k=1}^{n}k+{\sum }_{k=1}^{n}1$

$=2\cdot \frac{n(n+1)(2n+1)}{6}+3\cdot \frac{n(n+1)}{2}+n$

$=\frac{n(n+1)(2n+1)}{3}+\frac{3n(n+1)}{2}+n$