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Question
Answer the following:
(a + bx) (1 − x)6 = 3 − 20x + cx2 + ..... then find a, b, c
Solution
Consider, (a + bx) (1 – x)6
= a(1 – x)6 + bx (1 – x)6
= a(1 – 6C1 x + 6C2 x2 – 6C3 x3 + …) + bx(1 – 6C1 x + 6C2 x2 – 6C3 x3 + …)
= a(1 – 6x + 15x2 – 20x3 + …) + bx(1 – 6x + 15x2 – 20x3 + …)
= a + (b – 6a)x + (15a – 6b)x2 + … …(i)
Since, (a + bx) (1 – x)6 = 3 – 20x + cx2 + …
∴ a + (b – 6a)x + (15a – 6b)x2 + …
= 3 – 20x + cx2 +… …[From (i)]
Equating both sides, we get
a = 3, b – 6a = – 20, 15a – 6b = c
∴ a = 3, b = –2 , c = 15(3) – 6(–2) = 57
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