Q5. Find the middle terms of (x − 1/x)9.
n = 9 is odd: the 5th and 6th terms. T5 = 9C4x5(−1/x)4 = 126x and T6 = 9C5x4(−1/x)5 = −126/x.
Q6. Show that (x + 1/x2)8 has no term independent of x.
Tr+1 = 8Crx8−rx−2r = 8Crx8−3r. 8 − 3r = 0 gives r = 8/3, which is not a whole number, so no term is independent of x.
Q7. Find the 4th term from the end of (x − 1/x)7.
There are 8 terms, so the 4th from the end is the (7 − 4 + 2)th = 5th from the start: T5 = 7C4x3(−1/x)4 = 35/x.
Q8. The coefficients of x2 and x3 in (3 + ax)9 are equal. Find a (a ≠ 0).
9C2 · 37a2 = 9C3 · 36a3. Divide by 36a2: 36 × 3 = 84a, so a = 108/84 = 9/7.
Q9. The coefficients of three consecutive terms of (1 + x)n are in the ratio 1 : 7 : 42. Find n.
Let them be nCr−1, nCr, nCr+1. Then (n − r + 1)/r = 7 and (n − r)/(r + 1) = 42/7 = 6, so n = 8r − 1 and n = 7r + 6. Hence r = 7 and n = 55.
Q10. What is the greatest coefficient in the expansion of (1 + x)11, and which terms have it?
n = 11 is odd, so the two middle terms, the 6th and 7th, have the greatest coefficient: 11C5 = 11C6 = 462.
Q11. Find the coefficient of x2 in (1 + x)3(1 − x)2.
(1 + x)3(1 − x)2 = (1 + x)(1 − x2)2 = (1 + x)(1 − 2x2 + x4). The x2 term comes only from 1 × (−2x2), so the coefficient is −2.