CP 4.3 - Sums of integers, squares and cubes | Edexcel A-Level Further Mathematics

Three standard sums can handle a much wider family of series than their names suggest. You will expand polynomial summands into integers, squares and cubes, then adjust the limits when a series starts part-way along the sequence.

The three standard sums and their limits

Can $24$ square layers contain exactly as many unit squares as one large square? Their total is

1^2+2^2+\cdots+24^2

Remarkably, this is $4900=70^2$ . You will confirm that exact match with one formula rather than twenty-four separate additions.

You already know

Sigma notation specifies a summand and its index limits. From A-level Mathematics, an arithmetic series has sum $n(a+l)/2$ , and algebraic brackets can be expanded before collecting terms. There are $N-m+1$ terms from index $m$ through index $N$ , including both endpoints.

The notation $\sum_{r=1}^{n}r^2$ means $1^2+2^2+\cdots+n^2$ , not $(1+2+\cdots+n)^2$ . Sigma applies to the whole summand once at each permitted integer value of the index.

Standard sums

For a positive integer $n$ :

\sum_{r=1}^{n}r=\frac{n(n+1)}{2}

\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6} \qquad \sum_{r=1}^{n}r^3=\frac{n^2(n+1)^2}{4}

The first follows from the arithmetic-series formula; the square and cube sums are also supplied explicitly.

In the formula booklet.

The sizes make sense: there are $n$ terms, and the largest square term is $n^2$ , so the square sum should be of order $n^3$ . The formula's leading term is $n^3/3$ . Similarly, the cube sum has leading term $n^4/4$ . A result of the wrong degree is a warning to inspect your substitution.

Constants require a different count. For example,

\sum_{r=1}^{n}5=5n \qquad \sum_{r=4}^{n}5=5(n-3)\quad(n\ge4)

These constant-sum results are not in the formula booklet: obtain them by counting terms. The index changes, but the value $5$ does not.

Key idea

Read the limits before choosing a formula. Standard sums start at $1$ ; a constant contributes its value times the number of terms.

This question checks the endpoint count that will matter whenever a polynomial includes a constant term.

Expanding a polynomial summand

Expand the expression inside sigma first, then sum each power separately; you cannot usually multiply two sums to obtain the sum of a product.

For instance, $r^2(r+2)=r^3+2r^2$ . At each index this is an identity, so adding it over all the indices preserves the identity. A fixed coefficient such as $2$ may be taken outside sigma.

Combining standard sums

For constants $A,B,C,D$ and positive integer $n$ :

\sum_{r=1}^{n}(Ar^3+Br^2+Cr+D) =A\sum_{r=1}^{n}r^3+B\sum_{r=1}^{n}r^2+C\sum_{r=1}^{n}r+Dn

The coefficients must be independent of the summation index.

Not in the formula booklet: learn it.

Keep the separate sums visible before substituting. That line shows which formula belongs to each term and makes a missing constant much easier to notice.

Worked example 1: a product giving squares and cubes

Question. Find $\sum_{r=1}^{n}r^2(r+2)$ , giving your answer as a single expression in $n$ .

Plan. Expand into a cube and a square, use their standard sums and collect over denominator $12$ .

\begin{aligned} \sum_{r=1}^{n}r^2(r+2) &=\sum_{r=1}^{n}r^3+2\sum_{r=1}^{n}r^2 &&\text{expand before summing}\\ &=\frac{n^2(n+1)^2}{4}+\frac{n(n+1)(2n+1)}{3} &&\text{use both standard sums}\\ &=\frac{n(n+1)}{12}\bigl[3n(n+1)+4(2n+1)\bigr] &&\text{take out a common factor}\\ &=\frac{n(n+1)(3n^2+11n+4)}{12} &&\text{collect the bracket} \end{aligned}

Why this step? The product is inside sigma, so its factors vary together at each index. Expanding first respects that connection; multiplying two independently summed factors would introduce unwanted cross-products.

Answer. $\displaystyle \frac{n(n+1)(3n^2+11n+4)}{12}$ .

Check. At $n=2$ , the actual sum is $1^2(3)+2^2(4)=19$ . The formula gives $2(3)(12+22+4)/12=19$ .

Worked example 2 changes one factor, from $r^2$ to $r+1$ . The expansion now includes a linear term and a constant, so all the endpoint counting from Part 1 comes into use.

Worked example 2: keeping the constant term

Question. Find $\sum_{r=1}^{n}(r+1)(r+2)$ in simplified form.

Plan. Expand to $r^2+3r+2$ , then use the square sum, integer sum and a count of $n$ constant terms.

\begin{aligned} \sum_{r=1}^{n}(r+1)(r+2) &=\sum_{r=1}^{n}r^2+3\sum_{r=1}^{n}r+2n &&\text{expand and split}\\ &=\frac{n(n+1)(2n+1)}{6}+\frac{3n(n+1)}{2}+2n &&\text{substitute standard sums}\\ &=\frac n6\bigl[(n+1)(2n+1)+9(n+1)+12\bigr] &&\text{use denominator }6\\ &=\frac{n(n^2+6n+11)}{3} &&\text{collect and simplify} \end{aligned}

Why this step? The constant $2$ occurs once for every index, so it contributes $2n$ . Replacing it by a lone $2$ would silently discard $n-1$ copies.

Answer. $\displaystyle \frac{n(n^2+6n+11)}{3}$ .

Check. At $n=2$ , $2(3)+3(4)=18$ , and the formula gives $2(4+12+11)/3=18$ .

These two derived sums are not in the formula booklet: build them from the standard formulae. The opening square-layer total uses a standard formula directly:

\sum_{r=1}^{24}r^2=\frac{24(25)(49)}{6}=4900=70^2

Twenty-four square layers really do contain enough unit squares to make a square of side $70$ , with none left over.

Watch out

Wrong: $\sum r^2(r+2)=(\sum r^2)(\sum(r+2))$ . At $n=2$ , that product is $5(7)=35$ , while the correct sum is $19$ . Right: expand the summand into powers before summing.

The following sum changes the second factor of Worked example 2. Expand both brackets so that the new constant is counted correctly.

Subtracting partial sums for a later start

A series starting above $1$ is the difference between two sums that do start at $1$ : remove every earlier term, stopping just before the new lower limit.

For example, $5^2+6^2+7^2+8^2$ is the sum of squares up to $8$ , minus the sum up to $4$ . Subtracting the sum up to $5$ would remove the first term you wanted to keep.

Changing the lower limit

If $S_N=\sum_{r=1}^{N}u_r$ , and $1\le m\le N$ , then

\sum_{r=m}^{N}u_r=S_N-S_{m-1}

Use $S_0=0$ when $m=1$ .

Not in the formula booklet: learn it.

When the limits contain $n$ , substitute the entire upper index into the formula. The $n$ inside a standard-sum formula is an input slot; for an upper limit $2n$ , every occurrence becomes $2n$ .

Worked example 3: squares from n plus one to twice n

Question. Find $\sum_{r=n+1}^{2n}r^2$ in factorised form, where $n$ is a positive integer.

Plan. Subtract the square sum up to $n$ from the square sum up to $2n$ .

\begin{aligned} \sum_{r=n+1}^{2n}r^2 &=\frac{(2n)(2n+1)(4n+1)}{6}-\frac{n(n+1)(2n+1)}{6} &&\text{upper partial sum minus earlier sum}\\ &=\frac{n(2n+1)}{6}\bigl[2(4n+1)-(n+1)\bigr] &&\text{factorise}\\ &=\frac{n(2n+1)(7n+1)}{6} &&\text{collect the bracket} \end{aligned}

Why this step? The first retained index is $n+1$ , so precisely the first $n$ terms must be removed. No re-indexing is needed.

Answer. $\displaystyle \frac{n(2n+1)(7n+1)}{6}$ .

Check. At $n=2$ , the sum is $3^2+4^2=25$ , and the formula gives $2(5)(15)/6=25$ .

Worked example 4 changes only the exponent, from squares to cubes. The same two endpoints are used, but the standard partial-sum formula changes.

Worked example 4: cubes over the same range

Question. Find $\sum_{r=n+1}^{2n}r^3$ in factorised form.

Plan. Use the difference of the cube sums up to $2n$ and $n$ , then take out $n^2/4$ .

\begin{aligned} \sum_{r=n+1}^{2n}r^3 &=\frac{(2n)^2(2n+1)^2}{4}-\frac{n^2(n+1)^2}{4} &&\text{subtract the partial sums}\\ &=\frac{n^2}{4}\bigl[4(2n+1)^2-(n+1)^2\bigr] &&\text{take out a common factor}\\ &=\frac{n^2}{4}(15n^2+14n+3) &&\text{expand the bracket}\\ &=\frac{n^2(3n+1)(5n+3)}{4} &&\text{factorise} \end{aligned}

Why this step? The squared factors in the cube-sum formula are squares of the upper index and its successor. At upper index $2n$ , they are $(2n)^2$ and $(2n+1)^2$ , not $(2n)^2$ and $(n+1)^2$ .

Answer. $\displaystyle \frac{n^2(3n+1)(5n+3)}{4}$ .

Check. At $n=2$ , $3^3+4^3=91$ , and the expression gives $4(7)(13)/4=91$ .

These range-specific results are not in the formula booklet: form them by subtracting the supplied standard sums. At a numerical later start, the same method gives $\sum_{r=5}^{8}r^2=204-30=174$ .

Watch out

Wrong: remove the partial sum up to the lower limit itself. Right: remove the sum ending one index before it. The number of retained terms is another check: indices $n+1$ through $2n$ contain exactly $n$ terms.

The next range starts and ends later. Decide which complete partial sums enclose it before doing the algebra.

Quick Recap

Use the supplied sums of integers, squares and cubes with their correct upper index.
Expand polynomial summands before splitting the sum; count a constant once per index.
For a sum from $m$ to $N$ , subtract the partial sum up to $m-1$ from the one up to $N$ .
Check a small value and the polynomial's degree before trusting a simplified expression.

The square-layer puzzle is exact: $1^2+\cdots+24^2=24(25)(49)/6=70^2$ .

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CP 4.3 - Sums of integers, squares and cubes

The three standard sums and their limits

Expanding a polynomial summand

Subtracting partial sums for a later start

Quick Recap

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