Expand ${{\left( 1+x \right)}^{8}}$ in ascending powers of $x$, up to and including the term in

Question

Expand ${{\left( 1+x \right)}^{8}}$ in ascending powers of $x$, up to and including the term in ${{x}^{3}}$. In the expansion of ${{(1+y+k{{y}^{2}})}^{8}}$ in ascending powers of $y$, the coefficient of ${{y}^{3}}$ is zero. Find the value of constant $k$. Given that $\left| x \right|<1$, expand $\frac{2x+3}{(x+1)(x+2)}$ in ascending powers of $x$, up to and including the term in ${{x}^{2}}$.

Accepted Answer

(a) $1+8x+28x^2+56x^3+\cdots$; $k=-1$. (b) $\frac32-\frac54x+\frac98x^2+\cdots$, valid for $|x|<1$.

Maclaurin Series and Power Series - Binomial Expansion as a Maclaurin Series

Maclaurin Series and Power Series - Binomial Expansion as a Maclaurin Series

Solution:

Solution locked