   1. Chapter 8 Class 11 Binomial Theorem
2. Serial order wise
3. Ex 8.1

Transcript

Ex8.1, 11 Find (a + b)4 – (a – b)4. Hence, evaluate ﷐﷐﷐﷮3﷯+﷐﷮2﷯﷯﷮4﷯– ﷐﷐﷐﷮3﷯−﷐﷮2﷯﷯﷮4﷯ . 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 + 4a3b + 6a2b2 + 4ab3 + b4 – a4 + 4a3b – 6a2b2 + 4ab3 – b4 = a4 – a4 + 6a2b2 – 6a2b2 + b4 – b4 + 4a3b + 4a3b + 4ab3 + 4ab3 = 0 + 0 + 0 + 8a3b + 8ab3 = 8 a3b + 8ab3 = 8ab ( a2 + b2 ) Thus, (a + b)4 – (a – b)4 = 8ab ( a2 + b2 ) We need to find ﷐﷐﷐﷮3﷯+﷐﷮2﷯﷯﷮4﷯– ﷐﷐﷐﷮3﷯−﷐﷮2﷯﷯﷮4﷯ Putting a = ﷐﷮3﷯ and b = ﷐﷮2﷯ ﷐﷐﷐﷮3﷯+﷐﷮2﷯﷯﷮4﷯– ﷐﷐﷐﷮3﷯−﷐﷮2﷯﷯﷮4﷯= 8﷐﷐﷮3﷯ ﷯﷐﷐﷮2﷯ ﷯ ﷐﷐﷐﷐﷮3﷯ ﷯﷮2﷯+ ﷐﷐﷐﷮2﷯ ﷯﷮2﷯﷯ = 8﷐﷮3﷯﷐﷮2﷯ ﷐3+2﷯ = 8﷐﷮3﷯﷐﷮2﷯ ﷐5﷯ = (8 × 5)﷐﷮3 ×2﷯ = 40 ﷐﷮6﷯

Ex 8.1 