4.4 Q-10

Exercise 4.4 — Question 10

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Key Formulas

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

${\sum }_{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$

${\sum }_{k=1}^n{k}^3={\left[\frac{n(n+1)}{2}\right]}^2$

Arithmetic-Geometric Series

An arithmetic-geometric sequence is formed by multiplying corresponding terms of an arithmetic sequence and a geometric sequence.

If the arithmetic sequence has first term $a$ and common difference $d$, and the geometric sequence has first term $1$ and common ratio $r$, then:

General term: $T_n=[a+(n-1)d]r^{n-1}$

Sum to $n$ terms (using the multiply-and-subtract method): $S_n=\frac{a-[a+(n-1)d]r^n}{1-r}+\frac{dr(1-r^{n-1})}{(1-r{)}^2},r\ne 1$

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

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Solution

(Insert the worked solution for Q-10 here, applying the appropriate formula from above.)

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