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Misc 5 - Chapter 8 Class 11 Binomial CBSE NCERT - Miscellaneous

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise
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Misc 5 Evaluate (﷐﷮3﷯ + ﷐﷮2﷯ )6 – (﷐﷮3﷯ – ﷐﷮2﷯ )6 . First we find (a + b)6 – (a – b)6 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)6 = = ﷐6!﷮0! ﷐6 −0﷯!﷯ a6 × 1 + ﷐6!﷮1 ! ﷐6 −1﷯ !﷯ a5 b + ﷐6!﷮2!﷐6 −2﷯!﷯ a4 b2 + ﷐6!﷮3 !﷐6 −3﷯!﷯ a3 b3 + ﷐6!﷮4 ! ﷐6 − 4 ﷯ !﷯a2 b4 + ﷐6!﷮5 !﷐ 6 −5﷯!﷯ ab5 + ﷐6!﷮6 ! ﷐6 −6﷯ !﷯ 1 × b6 = ﷐6!﷮1 × 6! ﷯ a6 + ﷐6!﷮1 × 5!﷯ a5 b + ﷐6!﷮2! × 4!﷯ a4 b2 + ﷐6!﷮3 ! 3!﷯ a3 b3 + ﷐6!﷮4 ! 2!﷯ a2 b4 + ﷐6!﷮5 ! ×1﷯ a b5 + ﷐6!﷮6 ! × 1﷯ b6 = ﷐6!﷮6!﷯ a6 + ﷐6 ×5!﷮5! ﷯ a5b + ﷐6 ×5 ×4!﷮2 × 4!﷯ a4 b2 + ﷐6 ×5 ×4 × 3!﷮3 × 2 × 1 × 3!﷯ a3 b3 + ﷐6 × 5 × 4!﷮2 × 1 × 4!﷯ a2 b4 + ﷐6 × 5!﷮1 × 5!﷯ ab5 + ﷐6!﷮6!﷯ b6 = a6 + 6a5b + (3 × 5)a4b2 + (5 × 4)a3b3 + (3 × 5)a2b4 + 6ab5 + b6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 Thus, (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 For (a – b)6 Replace b by (–b) in (1) (a + (–b)) 6 = a6 + 6a5 (–b) + 15a4 (– b)2 + 20a3 (– b)3 + 15a2 (– b)4 + 6a(– b)5 + (– b)6 (a – b)6 = a6 – 6a5 b + 15a4 b2 – 20a3 b3 + 15a2 b4 – 6ab5 + b6 Now, (a + b)6 – (a – b)6 =(a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6) – (a6 – 6a5 b + 15a4 b2 – 20a3 b3 + 15a2 b4 – 6ab5 + b6 ) = a6 – a6 + 6a5 b + 6a5b + 15a4 b2 – 15a4 b2 + 20a3 b3 + 20a3 b3 + 15a2 b4 – 15a2 b4 + 6ab5 + 6ab5 + b6 – b6 = 0 + 12a5b + 0 + 40a3b3 + 0 + 12ab5 + 0 = 12a5b + 40a3b3 + 12ab5 = 4ab (3a4 + 10a2b2 + 3b4) Thus, (a + b)6 – (a – b)6 = 4ab (3a4 + 10a2b2 + 3b4) We need to find (﷐﷮3﷯ + ﷐﷮2﷯)6 – (﷐﷮3﷯ – ﷐﷮2﷯)6 Putting a = ﷐﷮3﷯ and b =﷐﷮2﷯ (﷐﷮3﷯ + ﷐﷮2﷯)6 – (﷐﷮3﷯ – ﷐﷮2﷯)6 = 4﷐﷐﷮3﷯ ﷯﷐﷐﷮2﷯ ﷯ ﷐3﷐﷐﷐﷮3﷯ ﷯﷮4﷯+10﷐﷐﷐﷮3﷯ ﷯﷮2﷯﷐﷐﷐﷮2﷯ ﷯﷮2﷯+3﷐﷐﷐﷮2﷯ ﷯﷮4﷯﷯ = 4﷐﷐﷮3×2﷯ ﷯ ﷐3﷐﷐﷐﷮3﷯ ﷯﷮4﷯+10﷐﷐﷐﷮3﷯ ﷯﷮2﷯﷐﷐﷐﷮2﷯ ﷯﷮2﷯+3﷐﷐﷐﷮2﷯ ﷯﷮4﷯﷯ = 4﷐﷐﷮6﷯ ﷯ ﷐3﷐﷐﷐﷮3﷯ ﷯﷮4﷯+10﷐﷐﷐﷮3﷯ ﷯﷮2﷯﷐﷐﷐﷮2﷯ ﷯﷮2﷯+3﷐﷐﷐﷮2﷯ ﷯﷮4﷯﷯ = 4﷐﷐﷮6﷯ ﷯ ﷐3﷐﷐3﷯﷮﷐1﷮2﷯ × 4﷯+10﷐﷐3﷯﷮﷐1﷮2﷯ × 2﷯﷐﷐2﷯﷮﷐1﷮2﷯ × 2﷯+3﷐﷐2﷯﷮﷐1﷮2﷯ × 4﷯﷯ = 4﷐﷐﷮6﷯ ﷯ ﷐3﷐﷐3﷯﷮2﷯+10﷐﷐3﷯﷮1﷯﷐﷐2﷯﷮1﷯+3﷐﷐2﷯﷮2﷯﷯ = 4﷐﷐﷮6﷯ ﷯ ﷐27+60+12﷯ = 4﷐﷐﷮6﷯ ﷯ ﷐99﷯ = 396﷐﷐﷮6﷯ ﷯ Thus, (﷐﷮3﷯ + ﷐﷮2﷯)6 – (﷐﷮3﷯ – ﷐﷮2﷯)6 = 396﷐﷐﷮6﷯ ﷯

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