Ex 8.2,10 - Chapter 8 Class 11 Binomial Theorem (Deleted)
Last updated at Jan. 29, 2020 by Teachoo
Last updated at Jan. 29, 2020 by Teachoo
Transcript
Ex 8.2, 10 The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r. Finding (r – 1)th term , rth & (r + 1)th term of (x + 1)n Writing (x + 1)n as (1 + x)n We know that general term of expansion (a + b)n is Tr+1 = nCr an – rbr For (1 + x)n , Putting a = 1 , b = x Tr+1 = nCr 1n – r xr Tr+1 = nCr xr ∴ Coefficient of (r + 1)th term = nCr For rth term of (1 + x)n Replacing r with r – 1 in (1) Tr – 1 + 1 = nCr – 1 xr – 1 Tr = nCr – 1 xr – 1 ∴ Coefficient of (r)th term = nCr – 1 For (r – 1)th term of (1 + x)n Replacing r with r – 2 in (1) Tr – 2 + 1 = nCr – 2 xr – 2 Tr – 1 = nCr – 2 xr – 2 ∴ Coefficient of (r – 1)th term = nCr-2 For rth term of (1 + x)n Replacing r with r – 1 in (1) Tr – 1 + 1 = nCr – 1 xr – 1 Tr = nCr – 1 xr – 1 ∴ Coefficient of (r)th term = nCr – 1 For (r – 1)th term of (1 + x)n Replacing r with r – 2 in (1) Tr – 2 + 1 = nCr – 2 xr – 2 Tr – 1 = nCr – 2 xr – 2 ∴ Coefficient of (r – 1)th term = nCr-2 Since the coefficient of (r – 1)th, rth and (r + 1)th terms are in ratio 1 : 3 : 5 (𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 〖(𝒓 − 𝟏)〗^𝒕𝒉 𝒕𝒆𝒓𝒎)/(𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒓^𝒕𝒉 𝒕𝒆𝒓𝒎) = 𝟏/𝟑 〖𝑛𝐶〗_(𝑟 − 2)/〖𝑛𝐶〗_(𝑟 − 1) = 1/3 (𝑛!/((𝑟 − 2)![𝑛 − (𝑟 − 2)]!))/(𝑛!/(𝑟 − 1)!(𝑛 − (𝑟 − 1))!) = 1/3 𝑛!/((𝑟. −2)![𝑛 − (𝑟 − 2)]!) × ((𝑟 − 1)![𝑛 − (𝑟 − 1)]!)/𝑛! = 1/3 (𝑛!(𝑟 − 1)(𝑟 − 2)![𝑛 − (𝑟 − 1)]!)/(𝑛!(𝑟 − 2)![𝑛 − (𝑟 − 2)]!) = 1/3 (𝑟 − 1)(𝑛 − (𝑟 − 1))!/((𝑛 − (𝑟 − 2))!) = 1/3 (𝑟 − 1)(𝑛− 𝑟 + 1)!/((𝑛 − 𝑟 + 2)!) = 1/3 (𝑟 − 1)(𝑛− 𝑟 + 1)!/((𝑛 − 𝑟 + 2)(𝑛 − 𝑟 + 2 −1)!) = 1/3 (𝑟 − 1)(𝑛− 𝑟 + 1)!/((𝑛 − 𝑟 + 2)(𝑛 − 𝑟 +1)!) = 1/3 ((𝑟 − 1))/((𝑛 − 𝑟 + 2) ) = 1/3 3(r – 1) = n – r + 2 3r – 3 = n + 2 – r n + 2 – r – 3r + 3 = 0 n – 4r + 5 = 0 Also (𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒓^𝒕𝒉 𝒕𝒆𝒓𝒎)/(𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 〖(𝒓 + 𝟏)〗^𝒕𝒉 𝒕𝒆𝒓𝒎) = 𝟑/𝟓 〖𝑛𝐶〗_(𝑟 − 1)/〖𝑛𝐶〗_𝑟 = 3/5 (𝑛!/((𝑟 − 1)![𝑛 − (𝑟 − 1)]!))/(𝑛!/(𝑟! (𝑛 − 𝑟)!)) = 3/5 𝑛!/((𝑟 − 1)!(𝑛 − 𝑟 + 1)!) × (𝑟! (𝑛 − 𝑟)! )/𝑛! = 3/5 (𝑛! × 𝑟 × (𝑟 − 1)!(𝑛 − 𝑟)!)/(𝑛!(𝑟 − 1)! (𝑛 − 𝑟 + 1)!) = 3/5 𝑟(𝑛 − 𝑟)!/((𝑛 − 𝑟 +1)!) = 3/5 (𝑟 (𝑛 − 𝑟)!)/((𝑛 − 𝑟 +1) (𝑛 − 𝑟)!) = 3/5 𝑟/(𝑛 − 𝑟 + 1) = 3/5 5r = 3 (n – r + 1) 5r = 3n – 3r + 3 0 = 3n – 3r + 3 – 5r 0 = 3n – 8r + 3 3n – 8r + 3 = 0 Now our equations are n – 4r + 5 = 0 & 3n – 8r + 3 = 0 From (1) n – 4r + 5 = 0 n = 4r – 5 Putting n = 4r – 5 in (2) 3n – 8r + 3 = 0 3(4r – 5) – 8r + 3 = 0 12r – 15 – 8r + 3 = 0 12r – 8r – 15 + 3 = 0 4r – 12 = 0 4r = 12 r = 12/4 r = 3 Putting r = 3 in n = 4r – 5 n = 4(3) – 5 n = 12 – 5 n = 7 Hence, n = 7 & r = 3
Ex 8.2
Ex 8.2, 2 Important Deleted for CBSE Board 2022 Exams
Ex 8.2, 3 Deleted for CBSE Board 2022 Exams
Ex 8.2, 4 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,5 Deleted for CBSE Board 2022 Exams
Ex 8.2 6 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,7 Deleted for CBSE Board 2022 Exams
Ex 8.2,8 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,9 Deleted for CBSE Board 2022 Exams
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Ex 8.2,11 Important Deleted for CBSE Board 2022 Exams
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Ex 8.2
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