Example 8 - The second, third, fourth terms in (x + a)n - Coefficient

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise

Transcript

Example 8 The second, third and fourth terms in the binomial expansion (x + a)n are 240, 720 and 1080, respectively. Find x, a and n. We know that general term of (a + b)n is Tr+1 = nCr (a)n-r . (b)r Given that second term of (x + a)n is 240 i.e. T2 = 240 T1+1 = 240 Putting r = 1 , a = x & b = a T1+1 = nCr (x)n-1 (a)1 T2 = nCr xn-1a 240 = nCr xn-1a Similarly, Third term of (x + a)n is 720 i.e. T3 = 720 T2+1 = 720 Putting r = 2 , a = x &b = a in (1) T2+1 = nC2 (x)n-2.(b)2 720 = nC2 xn-2.a2 Also , fourth term of (x + a)n is 1080 i.e. T4 = 1080 T3+1 = 1080 Putting r = 3 , a = x & b = a in (1) T3+1 = nC3 xn-3 . a3 1080 = nC3 xn-3 . a3 Dividing (3) by (2) ﷐720﷮240﷯ = ﷐𝑛𝐶2 ﷐𝑥﷮𝑛−2﷯﷐𝑎﷮2﷯﷮𝑛𝐶1 ﷐𝑥﷮𝑛−1 ﷯𝑎﷯ 3 = ﷐𝑛𝐶2 ﷮𝑛𝐶1 ﷯ × ﷐ ﷐𝑥﷮𝑛−2﷯ ﷮﷐𝑥﷮𝑛−1 ﷯﷯ × ﷐ ﷐𝑎﷮2﷯﷮𝑎﷯ 3 = ﷐﷐𝑛!﷮2!﷐𝑛−2﷯!﷯﷮﷐𝑛!﷮1!﷐𝑛−1﷯!﷯﷯ × xn – 2 – (n – 1) × a1 3 = ﷐𝑛!﷮2!﷐𝑛−2﷯!﷯×﷐1!﷐𝑛−1﷯! ﷮𝑛!﷯ × xn – 2 – n + 1 × a 3 = ﷐𝑛!﷮2﷐𝑛−2﷯!﷯×﷐1﷐𝑛−1﷯(𝑛−2)! ﷮𝑛!﷯ × x–1 × a 3 = ﷐﷐𝑛−1﷯ ﷮2﷯ × ﷐𝑎﷮𝑥﷯ 3 × 2 = (n – 1) × ﷐𝑎﷮𝑥﷯ 6 = (n – 1) × ﷐𝑎﷮𝑥﷯ ﷐6﷮𝑛−1﷯ = ﷐𝑎﷮𝑥﷯ Now Dividing (4) by (3) ﷐1080﷮720﷯ = ﷐𝑛𝐶3 ﷐𝑥﷮𝑛−3﷯﷐𝑎﷮3﷯﷮𝑛𝐶2 ﷐﷐𝑥﷮𝑛−2 ﷯𝑎﷮2﷯﷯ ﷐3﷮2﷯ = ﷐𝑛𝐶3 ﷮𝑛𝐶2 ﷯ × ﷐ ﷐𝑥﷮𝑛−3﷯ ﷮﷐𝑥﷮𝑛−2 ﷯﷯ × ﷐﷐𝑎﷮3﷯﷮﷐𝑎﷮2﷯﷯ ﷐3﷮2﷯ = ﷐﷐𝑛!﷮3!﷐𝑛−3﷯!﷯﷮﷐𝑛!﷮2!﷐𝑛−2﷯!﷯﷯ × xn – 3 – (n – 2) × a3 – 2 ﷐3﷮2﷯ = ﷐𝑛!﷮3!﷐𝑛 −3﷯!﷯ × ﷐2!﷐𝑛−2﷯!﷮𝑛!﷯ × x-1 × a1 ﷐3﷮2﷯ = ﷐𝑛!﷮3(2)!﷐𝑛 −3﷯!﷯ × ﷐2!﷐𝑛−2﷯(𝑛−3)!﷮𝑛!﷯ × ﷐𝑎﷮𝑥﷯ ﷐3﷮2﷯ = ﷐(𝑛 −2)﷮3﷯ × ﷐𝑎﷮𝑥﷯ ﷐9﷮2(𝑛 −2)﷯ = ﷐𝑎﷮𝑥﷯ Now our equations are ﷐6﷮𝑛 − 1﷯ = ﷐𝑎﷮𝑥﷯ …(A) ﷐9﷮2(𝑛 − 2)﷯ = ﷐𝑎﷮𝑥﷯ …(B) Equating (A) & (B) ﷐9﷮2(𝑛 −2)﷯ = ﷐6﷮𝑛 −1﷯ 9 × (n – 1) = 6 × 2(n – 2) 9n – 9 = 12n – 24 –9 + 24 = 12n – 9n 15 = 3n ﷐15﷮3﷯ = n 5 = n n = 5 Putting n = 5 in (A) ﷐6﷮𝑛 − 1﷯ = ﷐𝑎﷮𝑥﷯ ﷐6﷮5 − 1﷯ = ﷐𝑎﷮𝑥﷯ ﷐6﷮4﷯ = ﷐𝑎﷮𝑥﷯ ﷐3﷮2﷯ = ﷐𝑎﷮𝑥﷯ 3x = 2a ﷐﷐3﷮2﷯﷯x = a a = ﷐﷐3﷮2﷯﷯x Putting n = 5,a = ﷐﷐3﷮2﷯﷯x in (2) 240 = nC1 xn – 1 . ﷐﷐3﷮2﷯﷯x 240 = 5C1 x5-1 . ﷐﷐3﷮2﷯﷯x 240 = ﷐5!﷮1 × 4!﷯ x4 . ﷐﷐3﷮2﷯﷯x 240 = ﷐5!﷮1 × 4!﷯ x4 . ﷐﷐3﷮2﷯﷯x 240 = ﷐5 × 4!﷮4﷯ x4 . ﷐﷐3﷮2﷯﷯x 240 = 5x4﷐﷐3﷮2﷯﷯x 240 = ﷐5 × 3﷮2﷯ × x5 ﷐240 × 2﷮5 × 3﷯ = x5 32 = x5 x5 = 32 x5 = (2)5 x = 2 Putting x = 2 in ﷐𝑎﷮𝑥﷯ = ﷐3﷮2﷯ ﷐𝑎﷮2﷯ = ﷐3﷮2﷯ a = 3 Hence, x = 2, a = 3, n = 5

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