find the 58th term in the expansion of (2x-3y)^100
we will need this formula
$\displaystyle {n\choose k}=\frac{n!}{k!(n-k)!}$
From the Binomial Therom this will generate the k+1 th term.
$\displaystyle {{n}\choose {k}}a^{n-k}b^k$
so we need n=100 k=57
$\displaystyle a=2x$ and $\displaystyle b=-3y$
evaluating at the above values gives
$\displaystyle {{n}\choose {k}}a^{n-k}b^k=\frac{100!}{57!(100-57)!}(2x)^{100-57}(-3y)^{57}$
I hope this helps