Ex 8.2,12

Transcript

Ex 8.2,12 Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6. We know that General term of expansion (a + b)n is Tr+1 = nCr an–r br General term of (1 + x)m is Putting n = m, a = 1, b = x Tr + 1 = mCr (1)m-r . (x)r = mCr (1) . (x)r = mCr (x)r Thus, Tr + 1 = mCr x We need coefficient of x2 So put r = 2 Tr + 2 = mC2 x2 Coefficient of x2 is mC2 It is given that coefficient of x2 is 6   i.e. mC2 = 6 ﷐m!﷮2!﷐m−2﷯!﷯ = 6 ﷐m(𝑚−1)(𝑚−2)!﷮2!﷐m−2﷯!﷯ = 6 ﷐𝑚 (𝑚−1)﷮2﷯ = 6 m (m–1) = 12 m2 – m = 12 m2 – m – 12 = 0 m2 – 4m + 3m – 12 = 0 m(m–4) + 3 (m–4) = 0 (m + 3) (m – 4) = 0 m + 3 =0 or m – 4 = 0 m = –3 or m = 4 But we need positive value of m Hence, m = 4