Example 16 - Chapter 8 Class 11 Binomial Theorem (Deleted)
Last updated at Jan. 29, 2020 by Teachoo
Last updated at Jan. 29, 2020 by Teachoo
Transcript
Example 16 The sum of the coefficients of the first three terms in the expansion of (x โ 3/x2)m , x โ 0, m being a natural number is 559. Find the term of the expansion containing x3. We know that General term of expansion (a + b)n is Tr + 1 = nCr an โ r br For (x โ ๐/๐๐)m Putting n = m , a = x , b = (โ3)/๐ฅ^2 Tr+1 = mCr xm โ r ((โ3)/๐ฅ^2 )^๐ Now we need to find first 3 terms i.e. T1 , T2 and T3 i.e. T0 + 1 , T1 + 1 and T2 + 1 For T1 T0 +1 = mC0 x mโ0 ((โ3)/๐ฅ2)^0 = mC0 xm ร 1 = mC0 xm = 1 (xm) โด Coefficient of first term = 1 For T2 T1+1 = mC1 x mโ1 ((โ3)/๐ฅ2)^1 = mC1 (x) mโ1 ร (-3) ร 1/๐ฅ2 = mC1 (x) mโ1 ร (-3) ร xโ2 = โ3 mC1 . Xm โ 1 โ 2 = โ3 mC1 .xm โ 3 โด Coefficient of second term = โ3 mC1 For T3 T2+1 = mC2 (x) mโ2 ((โ3)/๐ฅ2)^2 = mC2 (x) m โ 2 (โ3)2 (1/๐ฅ2)^2 = mC2 (x) m โ 2 ร (9) ร xโ4 = mC2 . 9 .xm โ 2 โ 4 = 9 mC2 .xm โ 6 Coefficient of third term = 9 mC2 Given that Sum of coefficients of first three terms = 559 1 + ( โ3 mC1 ) + 9 mC2 = 559 1 โ 3 ร ๐!/1!(๐ โ 1)! + 9 ร ๐!/2!(๐ โ 2)! = 559 1 โ 3 ร ๐(๐ โ 1)!/( (๐ โ 1)!) + 9 ร (๐ (๐ โ 1)(๐ โ 2)!)/(2 ร (๐ โ 2)!) = 559 1 โ 3m + 9/2 m(m โ 1) = 559 (2 โ 2(3๐) + 9๐ (๐ โ 1))/2 = 559 (2 โ 6๐ + 9๐ (๐ โ 1))/2 = 559 2 โ 6m + 9m2 โ 9m = 559 ร 2 2 โ 15m + 9m2 = 1118 9m2 โ 15m + 2 โ 1118 = 0 9m2 โ 15m โ 1116 = 0 3(3m2 โ 5m โ 372) = 0 3m2 โ 5m โ 372 = 0 3m2 โ 36m + 31m โ 372 = 0 3m (m โ 12) + 31 (m โ 12) = 0 (3m + 31) (m โ 12) = 0 Hence, Since m is natural number. Hence, m = 12 Now, we have to find the term of the expansion containing x3 3m + 31 = 0 3m = โ31 m = (โ31)/3 m โ 12 = 0 m = 12 Finding general term Tr+1 = mCr xm โ r ((โ3)/๐ฅ2)^๐ = 12Cr x12 โ r ((โ3)/๐ฅ2)^๐ = 12Cr (x)12 โ r ( โ3)r (1/๐ฅ2)^๐ = 12Cr (x)12 โ r ( โ3)r x-2r = 12Cr . (โ3)r .x12โ r โ 2r = 12Cr . (โ3)r .x12 โ 3r We need the term containing x3 โด x3 = x12 โ 3r (Putting m = 12) Comparing powers 3 = 12 โ 3r 3r = 12 โ 3 3r = 9 r = 9/3 r = 3 Hence, the term containing x3 Tr +1 = 12Cr . (โ3)r . x12 โ 3r T3+1 = 12C3 (โ3)3 . x12 โ 3(3) = 12!/3!(12 โ13)! (โ3)3 . x3 = 12!/(3! 9!) (โ27).x3 = (12 ร 11 ร 10 ร 9!)/(3 ร 2 ร 1 ร 9!) (โ27) x3 = (12 ร 11 ร 10 )/(3 ร 2 ) (โ27) x3 = (12 ร 11 ร 10 )/(3 ร 2 ) (โ27) x3 = โ 5940 x3 Hence the Required term T4 is โ5940 x3
Examples
Example 2 Important Deleted for CBSE Board 2022 Exams
Example 3 Important Deleted for CBSE Board 2022 Exams
Example 4 Deleted for CBSE Board 2022 Exams
Example 5 Important Deleted for CBSE Board 2022 Exams
Example 6 Important Deleted for CBSE Board 2022 Exams
Example 7 Deleted for CBSE Board 2022 Exams
Example 8 Important Deleted for CBSE Board 2022 Exams
Example 9 Deleted for CBSE Board 2022 Exams
Example 10 Important Deleted for CBSE Board 2022 Exams
Example 11 Important Deleted for CBSE Board 2022 Exams
Example 12 Deleted for CBSE Board 2022 Exams
Example 13 Important Deleted for CBSE Board 2022 Exams
Example 14 Important Deleted for CBSE Board 2022 Exams
Example 15 Important Deleted for CBSE Board 2022 Exams
Example 16 Deleted for CBSE Board 2022 Exams You are here
Example 17 Important Deleted for CBSE Board 2022 Exams
Examples
About the Author