Ps1 - Standard summation formulae | OCR A-Level Further Mathematics B (MEI)

Sum expressions involving integers, squares and cubes without adding every term. You will combine the standard formulae, handle a changed starting index, and prove the square and cube formulae when a question requires justification.

Combining the standard sums

What is $1^2+3^2+5^2+\cdots+199^2$ , without a hundred separate squarings?

The odd numbers are $2r-1$ , for $r=1, 2,\ldots,100$ . Squaring gives $(2r-1)^2=4r^2-4r+1$ . That turns the puzzle into three familiar types of sum. We will finish it after practising that conversion.

You already know

$\sum_{r=1}^{n}u_r$ adds the terms for integer indices $1$ through $n$ , inclusive. Split a sum across addition and take constant factors outside it. The sum of $n$ copies of $c$ is $cn$ , not $c$ .

The three formulae below all start at $1$ . Match both the expression and the limits before substituting.

Integers, squares and cubes

For every positive integer $n$ ,

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

The integer sum follows from the arithmetic-series formula in the booklet. The square and cube formulae are printed explicitly.

In the formula booklet.

The cube formula has a lovely consequence: the sum of the first $n$ cubes is the square of the sum of the first $n$ integers. For $n=4$ , $1+8+27+64=100=(1+2+3+4)^2$ . The nested-square picture makes the identity less mysterious.

[DIAGRAM: asset_name: Cubes inside a triangular-number square; asset_slug: cubes_triangular_square; caption: Successive square bands have areas 1 cubed, 2 cubed, 3 cubed and 4 cubed; description: Nested squares with side lengths 1, 3, 6 and 10 occupy a ten by ten square. The successive L-shaped bands have areas 1, 8, 27 and 64, labelled in a colour-matched legend. The full side is labelled 1+2+3+4=10 and the total area is 100.]
Successive square bands have areas 1 cubed, 2 cubed, 3 cubed and 4 cubed

Worked example 1: expanding before summing

Question. Find $\sum_{r=1}^{n}r(r+1)$ in factorised form.

Plan. Expand, apply the integer and square formulae, then factorise.

\begin{aligned} \sum_{r=1}^{n}r(r+1)&=\sum_{r=1}^{n}r^2+\sum_{r=1}^{n}r && \text{expand first} \\ &=\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2 && \text{standard sums} \\ &=\frac{n(n+1)}6(2n+1+3) && \text{common factor} \\ &=\frac{n(n+1)(n+2)}3 && \text{simplify} \end{aligned}

Why this step? Expansion exposes the two formulae already available, so there is no separate product-sum formula to memorise.

Answer. $\frac13n(n+1)(n+2)$ .

Check. At $n=2$ , the sum is $2+6=8$ , and the formula gives $2(3)(4)/3=8$ .

Change only the second factor from $r+1$ to $r+3$ . The coefficient of the integer sum changes.

Worked example 2: a changed linear factor

Question. Show that $\sum_{r=1}^{n}r(r+3)=\frac13n(n+1)(n+5)$ .

Plan. Expand the product, separate the sums and simplify to the given expression.

\begin{aligned} \sum_{r=1}^{n}r(r+3)&=\sum_{r=1}^{n}r^2+3\sum_{r=1}^{n}r && \text{expand} \\ &=\frac{n(n+1)(2n+1)}6+\frac{3n(n+1)}2 && \text{substitute} \\ &=\frac{n(n+1)}6(2n+1+9) && \text{common denominator} \\ &=\frac{n(n+1)(n+5)}3 && \text{factorise} \end{aligned}

Why this step? The factor $3$ multiplies the whole integer sum, since it occurs at every index.

Answer. The required expression follows.

Check. For $n=1$ , both sides equal $4$ .

A “show that” answer needs the expansion and formula substitutions. Quoting the given expression does not establish it.

Watch out

Wrong: $\sum(r^2+3r)=\sum r^2+3$ . Right: $\sum r^2+3\sum r$ . The $3r$ contribution occurs at every index.

Change the second factor once more and choose the same route.

Changed limits and shifted terms

A sum starting after $1$ is the full sum minus the unwanted beginning; a shifted expression is expanded before the standard formulae are used.

For example, summing from $r=4$ through $r=10$ includes seven terms, and removes the terms indexed $1, 2, 3$ . It does not remove the term indexed $4$ .

Subtracting an initial section

If $1\le m\le n$ are integers,

\sum_{r=m}^{n}u_r=\sum_{r=1}^{n}u_r-\sum_{r=1}^{m-1}u_r

When $m=1$ , the second sum is empty and equals zero.

Not in the formula booklet: learn it.

Write the last removed index explicitly to prevent an off-by-one error.

Worked example 3: a tail of squares

Question. Find $\sum_{r=6}^{12}r^2$ .

Plan. Subtract the first five squares from the first twelve squares.

\begin{aligned} \sum_{r=6}^{12}r^2&=\sum_{r=1}^{12}r^2-\sum_{r=1}^{5}r^2 && \text{remove indices }1\text{ to }5 \\ &=\frac{12(13)(25)}6-\frac{5(6)(11)}6 && \text{square formula} \\ &=650-55=595 && \text{evaluate} \end{aligned}

Why this step? Subtracting through $6$ would remove the first term you want to keep.

Answer. $595$ .

Check. $36+49+64+81+100+121+144=595$ .

Now change the starting index to $n+1$ , with the upper index $2n$ . The subtraction becomes symbolic.

Worked example 4: the second block of squares

Question. Show that $\sum_{r=n+1}^{2n}r^2=\frac16n(2n+1)(7n+1)$ , for positive integers $n$ .

Plan. Subtract the first $n$ squares from the first $2n$ squares.

\begin{aligned} \sum_{r=n+1}^{2n}r^2&=\frac{2n(2n+1)(4n+1)}6-\frac{n(n+1)(2n+1)}6 && \text{initial block removed} \\ &=\frac{n(2n+1)}6\bigl(2(4n+1)-(n+1)\bigr) && \text{factorise} \\ &=\frac{n(2n+1)(7n+1)}6 && \text{collect terms} \end{aligned}

Why this step? The square formula uses its upper index everywhere: replacing $n$ by $2n$ changes $2n+1$ to $4n+1$ too.

Answer. The given formula is established.

Check. At $n=2$ , $3^2+4^2=25=2(5)(15)/6$ .

For the opening odd squares, expansion now gives

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

Thus the hundred odd squares total $100(39999)/3=1\,333\,300$ . This derived identity is not printed separately in the booklet.

Watch out

Wrong: $\sum_{r=1}^{n}(r+1)^2$ is the first $n$ squares. Right: either expand $(r+1)^2$ , or sum the squares from $2$ to $n+1$ . The shift changes the terms.

Apply block subtraction with cubes instead of squares.

Proving a summation formula

A supplied formula can still need a proof: induction verifies the first sum and shows that adding the next term preserves the claimed form.

You already know

An induction proof needs a base case, an assumption for an arbitrary integer $k$ , an implication proving the case $k+1$ , and a conclusion. The next sum is the previous sum plus its next term.

Use the assumed expression for the first $k$ terms. Substituting $k+1$ into the target formula alone would just restate what you must prove.

Worked example 5: proving the square formula

Question. Prove by induction that $\sum_{r=1}^{n}r^2=\frac16n(n+1)(2n+1)$ for all positive integers $n$ .

Plan. Check $n=1$ , assume the sum through $k$ , then add $(k+1)^2$ .

For $n=1$ , both sides equal $1$ . Assume the result for an arbitrary integer $k\ge1$ . Then

\begin{aligned} \sum_{r=1}^{k+1}r^2&=\frac{k(k+1)(2k+1)}6+(k+1)^2 && \text{use the assumption} \\ &=\frac{k+1}{6}\bigl(k(2k+1)+6(k+1)\bigr) && \text{common factor} \\ &=\frac{(k+1)(2k^2+7k+6)}6 && \text{expand} \\ &=\frac{(k+1)(k+2)(2k+3)}6 && \text{factorise} \end{aligned}

Why this step? The last line is the proposed formula at $n=k+1$ , because $2(k+1)+1=2k+3$ .

Answer. The result holds for $1$ , and truth for $k$ implies truth for $k+1$ ; hence it holds for every positive integer $n$ , by induction.

Check. At $k=2$ , $5+9=14=3(4)(7)/6$ .

Change only the power from squares to cubes. The added term becomes $(k+1)^3$ .

Worked example 6: proving the cube formula

Question. Prove by induction that $\sum_{r=1}^{n}r^3=\frac14n^2(n+1)^2$ for positive integers $n$ .

Plan. Verify $n=1$ , assume the sum through $k$ , and add the next cube.

At $n=1$ , both sides equal $1$ . Assume the result for an arbitrary integer $k\ge1$ . Then

\begin{aligned} \sum_{r=1}^{k+1}r^3&=\frac{k^2(k+1)^2}{4}+(k+1)^3 && \text{use the assumption} \\ &=\frac{(k+1)^2}{4}\bigl(k^2+4k+4\bigr) && \text{common factor} \\ &=\frac{(k+1)^2(k+2)^2}{4} && \text{perfect square} \end{aligned}

Why this step? Factoring out $(k+1)^2$ leaves the exact square needed for the next case.

Answer. The result holds for $1$ , and truth for $k$ implies truth for $k+1$ ; therefore it holds for all positive integers, by induction.

Check. At $k=2$ , $9+27=36=3^2(4^2)/4$ .

These proofs explain the formulae printed in the booklet; the induction arguments themselves are not provided.

Watch out

Wrong: add $k^3$ to the sum through $k$ . Right: add $(k+1)^3$ . The term indexed $k$ is already included.

Keep the cube method but multiply each input by $2$ . The added term and the claimed total both change.

Quick Recap

Expand polynomial summands before applying the standard sums.
A constant is added once for each index.
For changed limits, subtract through the index immediately before the desired start.
Proving a supplied formula still needs an actual induction step and conclusion.

The hundred odd squares add to $1\,333\,300$ .

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Ps1 - Standard summation formulae

Combining the standard sums

Changed limits and shifted terms

Proving a summation formula

Quick Recap

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