4.8 Q-9

Exercise 4.8 — Question 9

This question involves evaluating series using sigma notation and the standard summation formulas for natural numbers, their squares, and cubes, as well as 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: ${\sum }_{k=1}^n{k}^3={\left[\frac{n(n+1)}{2}\right]}^2$

General term of an arithmetic-geometric sequence (first term $a$, common difference $d$, common ratio $r$): $T_n=[a+(n-1)d]r^{n-1}$

Sum to infinity of an arithmetic-geometric series (when $|r|<1$): $S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$

---

Worked Solution

For a series that can be expressed using sigma notation, break it into standard sums:

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

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

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

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

$=\frac{n(n+1)(2n+1+3)}{6}=\frac{n(n+1)(2n+4)}{6}$

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

---

Tips for Exam

• Always split a combined series into standard $\sum k$, $\sum k^2$, $\sum k^3$ components.
• For arithmetic-geometric series, use the subtraction method ($S-rS$) to derive the sum.
• The convergence condition for an arithmetic-geometric series is $|r|<1$.