Finding Terms from Sums | CAIE 9709 Pure Mathematics 1 Notes

Chapters

Pure Mathematics 1

Mathematics9709

Subtract the part already counted

$S_5$ includes the first five terms; $S_4$ includes the first four. Their difference leaves only the fifth term.

The shared part is u₁ + u₂ + u₃ + u₄.

S₅ =shared part+ u₅

S₄ =shared part

S₅ − S₄ = u₅

The shared first four terms cancel.

u_n=S_n-S_{n-1}

This works for any sequence, not just an AP. For $n=1$ , use $u_1=S_1$ . The empty sum is $S_0=0$ ; do not assume a supplied formula for positive $n$ is valid at zero.

Worked example

Given $S_n=3n^2+2n$ , find $u_n$ .

Replace every $n$ by $n-1$ to get the previous sum. Subtract the whole expression:

\begin{aligned}S_{n-1}={}&3(n-1)^2\\&+2(n-1)\\={}&3n^2-4n+1\end{aligned}

\begin{aligned}u_n={}&3n^2+2n\\&-(3n^2-4n+1)\end{aligned}

u_n=6n-1

The terms are 5, 11, 17, …, an AP with $a=5$ and $d=6$ . Check: $5+11+17=33=S_3$ .

The next terms are not the first terms

To add terms 6 through 10, remove the first five terms from the first ten:

u_6+\cdots+u_{10}=S_{10}-S_5

More generally, terms $p$ through $q$ , including both ends, sum to $S_q-S_{p-1}$ . Subtracting $S_p$ would remove the first term you wanted.

If “the first five terms sum to 40 and the next five sum to 90”, write $S_5=40$ and $S_{10}-S_5=90$ , hence $S_{10}=130$ .

Now you try

Your turnIndependent practice

A sequence has $S_n=2n^2+5n$ . Find its nth term and the sum of terms 5 through 12. Decide whether the sequence is arithmetic.

Show worked answer

\begin{aligned}S_{n-1}&=2(n-1)^2+5(n-1)\\&=2n^2+n-3\end{aligned}

\begin{aligned}u_n={}&2n^2+5n\\&-(2n^2+n-3)\end{aligned}

u_n=4n+3

It is an AP with first term 7 and difference 4.

S_{12}-S_4=348-52=296

Check: eight terms run from 23 to 51. Their sum is $8(23+51)/2=296$ .