Sum

Answer the following:

Find the middle term (s) in the expansion of (x^{2}+ 2y^{2})^{7 }

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#### Solution

Here a = x^{2}, b = 2y^{2}, n = 7

Since n is odd

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

∴ Middle terms are t_{4} and t_{5}, for which r = 3 and r = 4 respectively.

We have, t_{r+1} = ^{n}C_{r} a^{n–r}. b^{r}

∴ t_{4} = ^{7}C_{3} (x^{2})^{4 }. (2y^{2})^{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`

= 280x^{8}y^{6}

and t_{5} = ^{7}C_{4} (x^{2})^{3 }. (2y^{2})^{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`

= 560x^{6}y^{8}

∴ Middle terms are 280x^{8}y^{6} and 560x^{6}y^{8}.

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

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