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

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 + ﷐﷮𝑎2−1﷯﷯﷮2﷯ + ﷐﷐a2 – ﷐﷮𝑎2−1﷯﷯﷮2﷯ ﷐﷐a2 + ﷐﷮𝑎2−1﷯﷯﷮2﷯ + ﷐﷐a2 – ﷐﷮𝑎2−1﷯﷯﷮2﷯ = 2﷐(a2)4 + 6(a2)2 ﷐﷐﷐﷮𝑎2−1﷯ ﷯﷮2﷯ + ﷐﷐﷐﷮𝑎2−1﷯ ﷯﷮4﷯﷯ = 2﷐a8 + 6a4 ﷐﷐𝑎2−1﷯﷮﷐1﷮2﷯ × 2﷯ +﷐﷐𝑎2−1﷯﷮﷐1﷮2﷯ × 4﷯﷯ = 2﷐a8 + 6a4 ﷐﷐𝑎2−1﷯﷮1﷯ +﷐﷐𝑎2−1﷯﷮2﷯﷯ = 2a8 + 12a4 (𝑎2−1) + 2 ( a4 + 1 – 2a2) = 2a8 + 12a6 – 12a4 + 2a4 + 2 – 4a2 = 2a8 + 12a6 – 10a4 – 4a2 + 2 So,﷐﷐a2 + ﷐﷮𝑎2−1﷯﷯﷮2﷯ + ﷐﷐a2 – ﷐﷮𝑎2−1﷯﷯﷮2﷯ = 2a8 + 12a6 – 10a4 – 4a2 + 2

Miscellaneous 