Discussion questions · Section 1 of 2

Using MF27, find the sum $\displaystyle\sum_{r=2}^{\infty}\frac{x^r}{r!}$, giving your answer in terms of $x$.

In the expansion of $(1+px-3x^2)^6$, the coefficient of $x^2$ is $42$. Find $p$ and the coefficient of $x^4$.

Express $f(x)=\dfrac{x^2+4x+3}{(2x+1)(x^2+1)}$ in partial fractions. Hence obtain its expansion in ascending powers of $x$ up to and including $x^3$. Find also the range of values of $x$ for which the expansion is valid.

Expand $(2+1/x)^{-2}$ in ascending powers of $x$, up to and including $x^5$. State the range of $x$ for which the expansion is valid.

The coefficient of $x^2$ in the series expansion of $\dfrac1{e^x(1+ax)}$ is $1/4$. Find the constant $a$. You may use standard results from MF27.

(i) Expand $\sqrt[3]{8+x}$ through $x^2$ and state the range of validity.

(ii) Put $x=1/27$ to approximate $\sqrt[3]{217}$ as $p/q$ in lowest terms, with $p,q\in\mathbb Z^+$.

(iii) We can also substitute $x=1/217$ into the expansion found in (i) to get an approximate value of $\sqrt[3]{217}$. Explain whether this approximation is better than the approximation in (ii).