Example 16 - Class 11

Transcript

Example 16 The sum of the coefficients of the first three terms in the expansion of (x – 3/x^2)^m , x ≠ 0, m being a natural number is 559. Find the term of the expansion containing x3. We know that general term of expansion (a + b)n is Tr+1 = nCr an–rbr for (x – ﷐3﷮𝑥2﷯)m Putting n = m , a = x , b = ﷐−3﷮𝑥2﷯ Tr+1 = mCr xm–r (﷐−3﷮𝑥2﷯)r Now we need to find first 3 terms i.e. T1 , T2 and T3 i.e. T0+1 , T1+1 and T2+1 Given that Sum of coefficients of first three terms = 559 1 + ( –3 mC1 ) + 9 mC2 = 559 1 – 3 × ﷐𝑚!﷮1!﷐𝑚−1﷯!﷯ + 9 × ﷐𝑚!﷮2!﷐ 𝑚−2﷯!﷯ = 559 1 – 3 × ﷐𝑚﷐𝑚−1﷯!﷮ ﷐𝑚−1﷯!﷯ + 9 × ﷐𝑚 ﷐𝑚−1﷯﷐ 𝑚−2﷯!﷮2 × ﷐ 𝑚−2﷯!﷯ = 559 1 – 3m + ﷐9﷮2﷯ m(m–1) = 559 ﷐2﷐1﷯−2(3𝑚) + 9𝑚 (𝑚−1)﷮2﷯ = 559 ﷐2 − 6𝑚 + 9𝑚 (𝑚−1)﷮2﷯ = 559 2 – 6m + 9m2 – 9m = 559 × 2 2 – 15m + 9m2 = 1118 9m2 – 15m + 2 – 1118 = 0 9m2 – 15m – 1116 = 0 3(3m2 – 5m – 372) = 0 3m2 – 5m – 372 = 0 3m2 – 36m + 31m – 372 = 0 3m (m – 12) + 31 (m – 12) = 0 (3m + 31) (m – 12) = 0 Hence, Since m is natural number. Hence m = 12 Now, we have to find the term of the expansion containing x3 Finding general term Tr+1 = mCr xm – r ﷐﷐﷐−3﷮𝑥2﷯﷯﷮𝑟﷯ = 12Cr x12 – r ﷐﷐﷐−3﷮𝑥2﷯﷯﷮𝑟﷯ = 12Cr (x)12 – r ( –3)r ﷐﷐﷐1﷮𝑥2﷯﷯﷮𝑟﷯ = 12Cr (x)12 – r ( –3)r x-2r = 12Cr . (–3)r .x12– r – 2r = 12Cr . (–3)r .x12 – 3r We need the term containing x3 ∴ x3 = x12 – 3r Comparing powers 3 = 12 – 3r 3r = 12 – 3 3r = 9 r = ﷐9﷮3﷯ r = 3 Hence, the term containing x3 Tr +1 = 12Cr . (–3)r . x12 – 3r T3+1 = 12C3 (–3)3 . x12 – 3(3) = ﷐12!﷮3!﷐12 −13﷯!﷯ (–3)3 . x3 = ﷐12!﷮3! 9!﷯ (–27).x3 = ﷐12 × 11 × 10 × 9!﷮3 × 2 ×1 × 9!﷯ (–27) x3 = ﷐12 × 11 × 10 ﷮3 × 2 ﷯ (–27) x3 = ﷐12 × 11 × 10 ﷮3 × 2 ﷯ (–27) x3 = – 5940 x3 Hence the Required term T4 is –5940 x3