Question

Question 2 Find the coefficient of a^5b^7 in (a – 2b)^12

Class X1 - Maths -Binomial Theorem Page 171

Answer 1

we know,
(r + 1)th term is the general term of the expansion of (x + y)ⁿ . And it is denoted by T_{r+1},
$T_{r+1}= ^nC_rx^{n-r}y^r$

The general term of (a -2b)¹² is
$T_{r+1}= ^{12}C_ra^{12-r}(-2)^rb^r\\= ^{12}C_ra^{12-r}b^r(-2)^r$
For coefficient of a⁵b⁷ put 12 -r = 5 ⇒r = 7
Now,

T₇₊₁ = ¹²C₇a¹²⁻⁷b⁷(-2)⁷
T₈ = ¹²C₇a⁵b⁷(-2)⁷

Hence, coefficient of a⁵b⁷ = ¹²C₇(-2)⁷
= {12 × 11 × 10 × 9 × 8 /5 × 4 × 3 × 2} (-2)⁷
= -11 × 9 × 8 × 128
= -101376

Answer 2

(r + 1)th term is the general term of the expansion of (x + y)ⁿ. And it is denoted by T_{r+1},  

The general term of (a -2b)¹² is

For coefficient of a⁵b⁷ put 12 -r = 5 ⇒r = 7

Now,

T₇₊₁ = ¹²C₇a¹²⁻⁷b⁷(-2)⁷

T₈ = ¹²C₇a⁵b⁷(-2)⁷

Hence, the coefficient of a⁵b⁷ = ¹²C₇(-2)⁷

= {12 × 11 × 10 × 9 × 8 /5 × 4 × 3 × 2} (-2)⁷

= -11 × 9 × 8 × 128

= -101376