Question

Find the middle term of (" title="( x^{2}+2) ^{20} 
" alt="( x^{2}+2) ^{20} 
" align="absmiddle" class="latex-formula">

Answer 1

So... when expanding a binomial, recall we get as many terms as the exponent plus 1, so, in this case will be 20 + 1, so 21 terms.

now, the middle guy in 21 terms, will be the 11th term.

using the binomial theorem, is simple to get the terms themselves, since all we're doing is, beginning at the highest with first element and going down by 1 each term, and with the second starting at 0 and going up, surely you already know that.

So, all that's left is what the dickens is the coefficient of each term in the expansion?

well, is "the current coefficient, times the exponent of the 1st element, divided all that by the exponent of the 2nd element on the next term", now, that's a mouthful but lemme post quick and explain a couple.

$\bf (x^2+2)^{20}\implies \begin{array}{llll} term&coefficient&value\\ -----&-----&-----\\ 1&&(x^2)^{20}(+2)^0\\ 2&+20&(x^2)^{19}(+2)^1\\ 3&+190&(x^2)^{18}(+2)^2\\ 4&+1140&(x^2)^{17}(+2)^3\\ 5&+4845&(x^2)^{16}(+2)^4\\ 6&+15504&(x^2)^{15}(+2)^5\\ 7&+38760&(x^2)^{14}(+2)^6\\ 8&+77520&(x^2)^{13}(+2)^7\\ 9&+125970&(x^2)^{12}(+2)^8\\ 10&+167960&(x^2)^{11}(+2)^9\\ 11&+184756&(x^2)^{10}(+2)^{10}\\ \end{array}$

so, how did we get the coefficient for say the 5th term, the coefficient of 4845?   well, is just 1140 * 17 / 4.

and how about the 6th term with 15504?  well 4845 * 16 / 5.

another example, say the 10th term's coefficient of 167960?  well is just 125970 * 12 / 9.

and now simply expand and combine the factors for the 11th term.