Combine standard finite sums

Use the natural-number, square and cube sums

The three special sums use positive integer upper bound $n$:

$$\sum_{r=1}^{n}r=\frac n2(n+1),\qquad \sum_{r=1}^{n}r^2=\frac n6(n+1)(2n+1),\qquad \sum_{r=1}^{n}r^3=\left[\frac n2(n+1)\right]^2.$$

The source examples take $n=4$: the respective sums are $10$, $30$ and $100$. Directly, $1+2+3+4=10$, $1^2+2^2+3^2+4^2=30$, and $1^3+2^3+3^3+4^3=100$.

The notes say that square/cube-sum formulas will be supplied when required in their examination context, whereas the natural-number formula is expected from the AP derivation. Treat that as guidance from these notes; follow the formulas and instructions supplied with the actual question. The QR callout points to the source’s visual-explanation resource for square and cube sums.

Combine polynomial and geometric sums without losing a term

Example 29(a) expands the product before using the square and natural-number formulas:

$$\begin{aligned} 1(1)+2(3)+\cdots+n(2n-1) &=\sum_{r=1}^{n}r(2r-1)\\ &=2\sum_{r=1}^{n}r^2-\sum_{r=1}^{n}r\\ &=\frac{2n(n+1)(2n+1)}6-\frac{n(n+1)}2\\ &=\frac{n(n+1)}6(4n+2-3)\\ &=\frac{n(n+1)(4n-1)}6. \end{aligned}$$

Part (b) starts at zero, so the geometric and constant groups have $n+1$ terms. Only the zero cube disappears:

$$\begin{aligned} \sum_{r=0}^{n}(ar^3+a^r-a) &=a\sum_{r=0}^{n}r^3+\sum_{r=0}^{n}a^r-\sum_{r=0}^{n}a\\ &=a\sum_{r=1}^{n}r^3+(1+a+a^2+\cdots+a^n)-(n+1)a\\ &=\frac{an^2(n+1)^2}{4}+\frac{1-a^{n+1}}{1-a}-(n+1)a\qquad(a\ne1). \end{aligned}$$

The middle group has first term 1, ratio $a$ and $n+1$ terms. This is a finite GP formula; it does not require $|a|<1$.

The source does not exclude $a=1$, so handle that missing case directly: the added and subtracted 1 cancel in each summand, giving $n^2(n+1)^2/4$.

At $a=0$, the power sum is interpreted as the polynomial $1+a+\cdots+a^n$ with constant term 1; it gives 1 without needing a separate analytic convention for $0^0$.

Although the source says “in terms of a”, the result also depends on the free upper bound $n$.

Part (c) is a range subtraction, not a sum beginning at 1:

$$\begin{aligned} (n+1)^2+(n+2)^2+\cdots+(2n)^2 &=\sum_{r=n+1}^{2n}r^2\\ &=\sum_{r=1}^{2n}r^2-\sum_{r=1}^{n}r^2\\ &=\frac{2n(2n+1)(4n+1)}6-\frac{n(n+1)(2n+1)}6\\ &=\frac n6(2n+1)[8n+2-(n+1)]\\ &=\frac n6(2n+1)(7n+1). \end{aligned}$$

Subtract through $n$ to retain the first requested square $(n+1)^2$.