I need to solve $\displaystyle (2-i)^5$ using Binomail Theorem which is $\displaystyle (z+v)^n = \sum_{j=o}^{n}\binom{n}{j} z^{n-j}v^j$ Now here z = 2, n = 5 and will v = -1 or $\displaystyle -i$ Thank you
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Originally Posted by flametag2 I need to solve $\displaystyle (2-i)^5$ using Binomail Theorem which is $\displaystyle (z+v)^n = \sum_{j=o}^{n}\binom{n}{j} z^{n-j}v^j$ Now here z = 2, n = 5 and will v = -1 or $\displaystyle -i$ Then why don't you do that? EXPAND $\displaystyle \sum_{j=0}^{5}\binom{5}{j} (2)^{n-j}(-i)^j$
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