# Find the middle terms in the expansion of (x4-1x3)11 - Mathematics and Statistics

Sum

Find the middle terms in the expansion of (x^4 - 1/x^3)^11

#### Solution

Here, a = x4 , b = (-1)/x^3, n = 11

Now, n is odd

∴ ("n" + 1)/2 = (11 + 1)/2 = 6,

("n" + 3)/2 = (11 + 3)/2 = 7

∴ Middle terms are t6 and t7, for which r = 5 and r = 6 respectively.

We have, tr+1 = nCr an–r .br

∴ t6 = ""^11"C"_5 (x^4)^6 ((-1)/x^3)^5

= (11!)/(5!6!) (x^24) xx ((-1)/x^15)

= (11 xx 10 xx 9 xx 8 xx 7)/(5 xx 4 xx 3 xx 2 xx 1) (-x^9)

= – 462x9

and t7 = ""^11"C"_6 (x^4)^5 ((-1)/x^3)^6

= (11!)/(6!5!) (x^4)^5 ((-1)/x^3)^6

= (11 xx 10 xx 9 xx 8 xx 7)/(5 xx 4 xx 3 xx 2 xx 1) (x^20) xx 1/x^18

= 462x2

∴ Middle terms are – 462x9 and 462x2

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
Exercise 4.3 | Q 4. (v) | Page 80