expand by the binomial theorem as far as the term $\displaystyle t^3$ , assuming that |3t| < 1; : $\displaystyle (1 - 3t)^-2 $ i know the answer is real easy i just forgotten how to do it thanks guys!.
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The binomial series is: $\displaystyle (1+x)^{m}=1+mx+\frac{m(m-1)}{2!}x^{2}+$$\displaystyle \frac{m(m-1)(m-2)}{3!}x^{3}+\frac{m(m-1)(m-2)(m-3)}{4!}x^{4}+...........$ Just sub in $\displaystyle m=-2, \;\ x=-3t$
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