Exercise
$\left(2-7a\right)^5$
Step-by-step Solution
Learn how to solve special products problems step by step online. Expand the expression (2-7a)^5. We can expand the expression \left(2-7a\right)^5 using Newton's binomial theorem, which is a formula that allow us to find the expanded form of a binomial raised to a positive integer n. The formula is as follows: \displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n. The number of terms resulting from the expansion always equals n + 1. The coefficients \left(\begin{matrix}n\\k\end{matrix}\right) are combinatorial numbers which correspond to the nth row of the Tartaglia triangle (or Pascal's triangle). In the formula, we can observe that the exponent of a decreases, from n to 0, while the exponent of b increases, from 0 to n. If one of the binomial terms is negative, the positive and negative signs alternate.. Calculate the power 2^{5}. Calculate the power 2^{4}. Calculate the power 2^{3}.
Expand the expression (2-7a)^5
Final answer to the exercise
$32-560a+80\left(-7a\right)^{2}+40\left(-7a\right)^{3}+10\left(-7a\right)^{4}+\left(-7a\right)^{5}$