Misc 9 - Expand using Binomial Theorem (1 + x/2 - 2/x)4 - Expansion

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

Transcript

Misc 9 Expand using Binomial Theorem ﷐﷐1 + ﷐x﷮2﷯−﷐2﷮x﷯﷯﷮4﷯, x ≠ 0 . 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 We need to find ﷐﷐1 + ﷐x﷮2﷯−﷐2﷮x﷯﷯﷮4﷯, Putting a = ﷐1 + ﷐𝑥﷮2﷯﷯ & b = ﷐﷐−2﷮𝑥﷯﷯ ﷐﷐﷐1 + ﷐x﷮2﷯﷯−﷐2﷮x﷯﷯﷮4﷯ = ﷐﷐1 + ﷐x﷮2﷯﷯﷮4﷯﷐﷐−2﷮x﷯﷯ + 4 ﷐1 +﷐x﷮2﷯﷯ ﷐﷐﷐2﷮x﷯﷯﷮3﷯ + 6 ﷐﷐1 +﷐x﷮2﷯﷯﷮2﷯ ﷐﷐﷐2﷮x﷯﷯﷮2﷯+ 4 ﷐1 +﷐x﷮2﷯﷯ ﷐﷐﷐2﷮x﷯﷯﷮3﷯ + ﷐﷐﷐−2﷮x﷯﷯﷮4﷯ = ﷐﷐1 + ﷐x﷮2﷯﷯﷮4﷯ – ﷐8﷮𝑥﷯ ﷐﷐1 +﷐x﷮2﷯﷯﷮3﷯ + ﷐24﷮𝑥2﷯ ﷐﷐1 +﷐x﷮4﷯﷯﷮2﷯ – ﷐32﷮𝑥2﷯ ﷐1 + ﷐x﷮2﷯﷯ + ﷐32﷮x4﷯ Now Solving ﷐﷐1 + ﷐x﷮2﷯﷯﷮4﷯ , ﷐﷐1 + ﷐x﷮2﷯﷯﷮3﷯ separately Solving ﷐﷐𝟏 + ﷐𝐱﷮𝟐﷯﷯﷮𝟒﷯ From (1) (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Putting a = 1 and b = ﷐𝑥﷮2﷯ ﷐﷐1 + ﷐x﷮2﷯﷯﷮4﷯ = (1)4 +(1)3 ﷐﷐x﷮2﷯﷯ + 6 (1)2 ﷐﷐﷐𝑥﷮2﷯﷯﷮2﷯ + 4 (1) ﷐﷐x﷮2﷯﷯ + ﷐﷐x﷮2﷯﷯ = 1 + 4 ﷐﷐x﷮2﷯﷯ + 4 ﷐﷐8﷮x﷯﷯ + 6 ﷐﷐﷐x﷮2﷯﷮2﷯﷯ + 4 ﷐﷐x3﷮8﷯﷯ + ﷐﷐x4﷮16﷯﷯ = 1 + 2x + ﷐3﷮2﷯ x2 + ﷐𝑥3﷮2﷯ + ﷐𝑥4﷮16﷯ Now Solving ﷐﷐𝟏 +﷐𝐱﷮𝟐﷯﷯﷮𝟑﷯ We know that (a + b )3 = a3 + 3a2b + 3ab + b3 Putting a = 1 & b = ﷐﷐﷐𝑥﷮2﷯﷯﷮3﷯ ﷐﷐1 +﷐x﷮2﷯﷯﷮3﷯ = (1)3 + 3 (1)2 + ﷐﷐x﷮2﷯﷯ + 3 (1) ﷐﷐﷐𝑥﷮2﷯﷯﷮2﷯ + ﷐﷐﷐x﷮2﷯﷯﷮3﷯ = 1 + 3﷐﷐x﷮2﷯﷯ + 3﷐﷐﷐𝑥﷮2﷯﷮2﷯﷯ + ﷐x3﷮8﷯ Now, ﷐﷐﷐1 + ﷐x﷮2﷯﷯−﷐2﷮x﷯﷯﷮4﷯ = ﷐﷐1 + ﷐x﷮2﷯﷯﷮4﷯ – ﷐8﷮𝑥﷯ ﷐﷐1 +﷐x﷮2﷯﷯﷮3﷯ + ﷐24﷮𝑥2﷯ ﷐﷐1 +﷐x﷮4﷯﷯﷮2﷯– ﷐32﷮𝑥2﷯ ﷐1 + ﷐x﷮2﷯﷯ + ﷐32﷮x4﷯ Putting value of ﷐﷐1 +﷐x﷮2﷯﷯﷮3﷯& ﷐﷐1 +﷐x﷮2﷯﷯﷮4﷯ = ﷐1+2𝑥+﷐3x2﷮2﷯+﷐x3﷮2﷯+﷐x4﷮16﷯﷯– ﷐8﷮x﷯ ﷐1 + ﷐3x﷮2﷯+﷐3x2﷮4﷯+﷐x3﷮8﷯﷯ + ﷐24﷮𝑥2﷯ ﷐1 +﷐﷐﷐x﷮2﷯﷯﷮2﷯+2 ﷐1﷯﷐﷐x﷮2﷯﷯﷯ – ﷐32﷮x3﷯ ﷐1 +﷐x﷮2﷯﷯ + ﷐16﷮x4﷯ = ﷐1+2𝑥+﷐3﷮2﷯ 𝑥2+﷐𝑥3﷮2﷯+﷐𝑥4﷮16﷯﷯ – ﷐﷐8﷮𝑥﷯ + 12 + 6𝑥 + 𝑥2﷯ + ﷐﷐24﷮x2﷯ +﷐24﷮x2﷯ + 6﷯ – ﷐﷐32﷮x3﷯ +﷐16﷮x2﷯﷯ + ﷐16﷮𝑥4﷯ = 1 + 2x + ﷐3﷮2﷯ x2 + ﷐𝑥3﷮2﷯ + ﷐𝑥4﷮16﷯ – ﷐8﷮𝑥﷯ + 12 + 6x + x2 + ﷐24﷮x2﷯ + ﷐24﷮x2﷯ + 6 – ﷐32﷮x3﷯ + ﷐16﷮x2﷯ + ﷐16﷮𝑥4﷯ = ﷐𝑥4﷮16﷯ + ﷐𝑥3﷮2﷯ + ﷐3﷮2﷯ x2 – x2 + 2x – 6 + 1 – ﷐8﷮𝑥﷯ + ﷐24﷮x﷯ + ﷐24﷮x2﷯ – ﷐16﷮x2﷯ – ﷐32﷮x3﷯ + ﷐16﷮𝑥4﷯ = ﷐𝑥4﷮16﷯ + ﷐𝑥3﷮2﷯ + ﷐𝑥2﷮2﷯ – 4x – 5 + ﷐16﷮𝑥﷯ + ﷐−32﷮𝑥2﷯ + ﷐16﷮𝑥4﷯ Thus, ﷐﷐1 + ﷐x﷮2﷯−﷐2﷮x﷯﷯﷮4﷯ = ﷐𝑥4﷮16﷯ + ﷐𝑥3﷮2﷯ + ﷐𝑥2﷮2﷯ – 4x – 5 + ﷐16﷮𝑥﷯ + ﷐−32﷮𝑥2﷯ + ﷐16﷮𝑥4﷯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.