# Ex 8.2,5 - Chapter 8 Class 11 Binomial Theorem (Deleted)

Last updated at Jan. 29, 2020 by Teachoo

Coefficient

Ex 8.2,5
Deleted for CBSE Board 2022 Exams
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Ex 8.2 6 Important Deleted for CBSE Board 2022 Exams

Example 5 Important Deleted for CBSE Board 2022 Exams

Ex 8.2,12 Deleted for CBSE Board 2022 Exams

Ex 8.2,9 Deleted for CBSE Board 2022 Exams

Ex 8.2,11 Important Deleted for CBSE Board 2022 Exams

Misc 2 Deleted for CBSE Board 2022 Exams

Example 17 Important Deleted for CBSE Board 2022 Exams

Misc 1 Important Deleted for CBSE Board 2022 Exams

Example 8 Important Deleted for CBSE Board 2022 Exams

Example 16 Deleted for CBSE Board 2022 Exams

Example 11 Important Deleted for CBSE Board 2022 Exams

Example 9 Deleted for CBSE Board 2022 Exams

Ex 8.2,10 Important Deleted for CBSE Board 2022 Exams

Example 10 Important Deleted for CBSE Board 2022 Exams

Example 15 Important Deleted for CBSE Board 2022 Exams

Example 13 Important Deleted for CBSE Board 2022 Exams

Misc 3 Important Deleted for CBSE Board 2022 Exams

Example 14 Important Deleted for CBSE Board 2022 Exams

Misc 8 Important Deleted for CBSE Board 2022 Exams

Ex 8.2, 5 Find the 4th term in the expansion of (x – 2y)12 . We know that General term of expansion (a + b)n is Tr+1 = nCr an–r br We need to find fourth term i.e. T4 = T3 + 1 of expansion (x – 12y)12 Putting r = 3 , a = x , b = –2y , n = 12 T3 + 1 = 12C3 (x)12 – 3 (–2y)3 = (12 !)/(3! (12 − 3)!) x9 (–2)3 (y)3 = 12!/(3! × 9!) x9 (–8) y3 = (12 × 11 × 10 × 9!)/(3 × 2 × 9 !) × (–8) × x9 y3 = (12 × 11 × 10 × (−8))/6 × x9 y3 = 2 × 11 × 10 × (–8) × x9 y3 = –220 × 8 x9 y3 = –1760 x9 y3