Lesson 7 of 10 · 6 min
Special cases
NCERT §7.2.2
A student at the stall tries (x − 2y)⁵ and gets every sign wrong. What happens to the theorem when one term of the binomial is negative, or is just 1?
The lesson in notes
In short
Putting b = −y: (x − y)ⁿ = nC0 xⁿ − nC1 xⁿ⁻¹y + nC2 xⁿ⁻²y² − … + (−1)ⁿ nCn yⁿ. The signs alternate, starting with +, because odd powers of −y are negative.
Example: (x − 2y)⁵ = x⁵ − 10x⁴y + 40x³y² − 80x²y³ + 80xy⁴ − 32y⁵.
Putting a = 1 and b = x: (1 + x)ⁿ = nC0 + nC1 x + nC2 x² + … + nCn xⁿ. Here the coefficient of xʳ is just nCr.
Putting x = 1 in that line gives the sum of a row: nC0 + nC1 + … + nCn = 2ⁿ. Row 6 adds to 64 = 2⁶.
Putting a = 1 and b = −x gives (1 − x)ⁿ = nC0 − nC1 x + nC2 x² − … + (−1)ⁿ nCn xⁿ, and x = 1 then gives nC0 − nC1 + nC2 − … + (−1)ⁿ nCn = 0.
The same trick finds the sum of the coefficients of any expansion: put the variable equal to 1. For (x − 3)⁴ the coefficients 1, −12, 54, −108, 81 add to (1 − 3)⁴ = 16.