# Misc 6 - Chapter 8 Class 11 Binomial Theorem

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 6 Find the value of (a2 + a2 −1)4 + (a2 – a2 −1)4 . First we find (a + b)4 + (a – b)4 and then solve We know that (a + b)n = nC0 an b0 + nC1 an – 1 b1 + nC2 an – 2 b2 + …. …. + nCn – 1 a1 bn – 1 + nCn a0 bn Hence, (a + b)4 = 4C0 a4 (b)0 + 4C1 a3 b1 + 4C2 a2 b2 + 4C3 a1b3 + 4C4 a0 b4 = 4!0! 4 − 0! a4 × 1 + 4!1×(4−1)! a2 b2 + 4!2!4 −2! ab3 + 4!4!(4 − 4)! 1 × b4 = 4!1 ×4! a4 + 4!1 ×3! a3 b + 4!2!(4 − 2)! a2 b2 + 4!3!4 −3! ab3 + 4!4! 0! b4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Hence, (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Replacing b with –b (a + (–b))4 = a4 + 4a3 (–b) + 6a2 (–b)2 + 4a (–b)3 (–b)4 (a – b)4 = a4 – 4a3 b + 6a2b2 – 4ab3 + b4 We need to find (a + b)4 + (a – b)4 (a + b)4 + (a – b)4 = (a4 + 4a3b + 6a2b2 + 4ab3 + b4 ) + (a4 – 4a3b + 6a2b2 – 4ab3 + b4 ) = a4 + a4 + 4ab3 – 4ab3 + 6a2b2 + 6a2b2 + 4a3b – 4a3b + b4 + b4 = 2a4 + 0 + 12 a2b2 + 0 + 2b4 = 2(a4 + 6 a2b2 + b4) Thus, (a + b)4 + (a – b)4 = 2(a4 + 6 a2b2 + b4) We need to find a2 + 𝑎2−12 + a2 – 𝑎2−12 a2 + 𝑎2−12 + a2 – 𝑎2−12 = 2(a2)4 + 6(a2)2 𝑎2−1 2 + 𝑎2−1 4 = 2a8 + 6a4 𝑎2−112 × 2 +𝑎2−112 × 4 = 2a8 + 6a4 𝑎2−11 +𝑎2−12 = 2a8 + 12a4 (𝑎2−1) + 2 ( a4 + 1 – 2a2) = 2a8 + 12a6 – 12a4 + 2a4 + 2 – 4a2 = 2a8 + 12a6 – 10a4 – 4a2 + 2 So,a2 + 𝑎2−12 + a2 – 𝑎2−12 = 2a8 + 12a6 – 10a4 – 4a2 + 2

About the Author

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.