The general term of an expansion, and the term counted from the end
List the expansion of (x + a)ⁿ as T1, T2, T3, … The term in place r + 1 uses nCr, so the general term is T(r+1) = nCr · xⁿ⁻ʳ · aʳ. The lower number of nCr is always one less than the order of the term: the third term has r = 2, the tenth has r = 9. If a is a fraction or holds x, keep it in brackets, for example (3/x)ʳ, and simplify afterwards. In (x − a)ⁿ the sign is that of (−a)ʳ, so terms with odd r are negative. To find the kth term from the end, count: the expansion has n + 1 terms, so the last term is number n + 1 from the start, the second from the end is number n, and in general the kth from the end is number n + 2 − k from the start. Another way to see it is to write the expansion backwards: the kth term from the end of (x + a)ⁿ is the kth term from the start of (a + x)ⁿ. Read the question carefully: a ‘term’ includes its power of x, the ‘coefficient’ is only the number in front, and the ‘power of x’ is only the exponent.
| Term | r | 5Cr | power of x | power of a |
|---|---|---|---|---|
| T1 | 0 | 1 | 5 | 0 |
| T2 | 1 | 5 | 4 | 1 |
| T3 | 2 | 10 | 3 | 2 |
| T4 | 3 | 10 | 2 | 3 |
| T5 | 4 | 5 | 1 | 4 |
| T6 | 5 | 1 | 0 | 5 |
Count from both ends. Write the 8 terms of (x + a)⁷ as boxes T1 to T8. Mark the 3rd from the start and the 3rd from the end, and say which numbers they have from the start.