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Question
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
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