Maharashtra State BoardHSC Science (General) 11th
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Answer the following: Find the middle term (s) in the expansion of (x2+ 2y2)7 - Mathematics and Statistics

Sum

Answer the following:

Find the middle term (s) in the expansion of (x2+ 2y2)

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Solution

Here a = x2, b = 2y2, n = 7

Since n is odd

∴ `("n" + 1)/2 = (7 + 1)/2` = 4, `("n" + 3)/2 = (7 + 3)/2` = 5

∴ Middle terms are t4 and t5, for which r = 3 and r = 4 respectively.

We have, tr+1 = nCr an–r. br 

∴ t4 = 7C3 (x2)4 . (2y2)3

= `(7!)/(3!4!) (x^8) (8y^6)`

= `(7 xx 6 xx 5 xx 4!)/(3 xx 2 xx 1 xx 4!) xx 8x^8y^6`

= 280x8y6

and t5 = 7C4 (x2)3 . (2y2)4

= `(7!)/(4!3!)(x^6).(16y^8)`

= `(7 xx 6 xx 5 xx 4!)/(4! xx 3 xx 2 xx 1) xx 16x^6y^8`

= 560x6y8

∴ Middle terms are 280x8y6 and 560x6y8.

Concept: Middle term(s) in the expansion of (a + b)n
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4 | Q II. (8) (iii) | Page 85
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