# Example 8 - Chapter 8 Class 11 Binomial Theorem

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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) 720240 = 𝑛𝐶2 𝑥𝑛−2𝑎2𝑛𝐶1 𝑥𝑛−1 𝑎 3 = 𝑛𝐶2 𝑛𝐶1 × 𝑥𝑛−2 𝑥𝑛−1 × 𝑎2𝑎 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!2𝑛−2!×1𝑛−1(𝑛−2)! 𝑛! × x–1 × a 3 = 𝑛−1 2 × 𝑎𝑥 3 × 2 = (n – 1) × 𝑎𝑥 6 = (n – 1) × 𝑎𝑥 6𝑛−1 = 𝑎𝑥 Now Dividing (4) by (3) 1080720 = 𝑛𝐶3 𝑥𝑛−3𝑎3𝑛𝐶2 𝑥𝑛−2 𝑎2 32 = 𝑛𝐶3 𝑛𝐶2 × 𝑥𝑛−3 𝑥𝑛−2 × 𝑎3𝑎2 32 = 𝑛!3!𝑛−3!𝑛!2!𝑛−2! × xn – 3 – (n – 2) × a3 – 2 32 = 𝑛!3!𝑛 −3! × 2!𝑛−2!𝑛! × x-1 × a1 32 = 𝑛!3(2)!𝑛 −3! × 2!𝑛−2(𝑛−3)!𝑛! × 𝑎𝑥 32 = (𝑛 −2)3 × 𝑎𝑥 92(𝑛 −2) = 𝑎𝑥 Now our equations are 6𝑛 − 1 = 𝑎𝑥 …(A) 92(𝑛 − 2) = 𝑎𝑥 …(B) Equating (A) & (B) 92(𝑛 −2) = 6𝑛 −1 9 × (n – 1) = 6 × 2(n – 2) 9n – 9 = 12n – 24 –9 + 24 = 12n – 9n 15 = 3n 153 = n 5 = n n = 5 Putting n = 5 in (A) 6𝑛 − 1 = 𝑎𝑥 65 − 1 = 𝑎𝑥 64 = 𝑎𝑥 32 = 𝑎𝑥 3x = 2a 32x = a a = 32x Putting n = 5,a = 32x in (2) 240 = nC1 xn – 1 . 32x 240 = 5C1 x5-1 . 32x 240 = 5!1 × 4! x4 . 32x 240 = 5!1 × 4! x4 . 32x 240 = 5 × 4!4 x4 . 32x 240 = 5x432x 240 = 5 × 32 × x5 240 × 25 × 3 = x5 32 = x5 x5 = 32 x5 = (2)5 x = 2 Putting x = 2 in 𝑎𝑥 = 32 𝑎2 = 32 a = 3 Hence, x = 2, a = 3, n = 5

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.