Solution:
(i) $1,-1 / 2,1 / 4,-1 / 8, \ldots$
It is known that, the sum of $\mathrm{GP}$ for infinity $=\mathrm{a} /(1-\mathrm{r})$
Given that,
$\mathrm{a}=1, \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=(-1 / 2) / 1=-1 / 2$
Substitute the values in
$\begin{array}{l}
\mathrm{a} /(1-\mathrm{r})=1 /(1-(-1 / 2)) \\
=1 /(1+1 / 2) \\
=1 /((2+1) / 2) \\
=1 /(3 / 2) \\
=2 / 3
\end{array}$
(ii) $\left(a^{2}-b^{2}\right),(a-b),(a-b) /(a+b), \ldots$ to $n$ terms
It is known that, the sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$
Given that,
$a=\left(a^{2}-b^{2}\right), r=t_{2} / t_{1}=(a-b) /\left(a^{2}-b^{2}\right)=(a-b) /(a-b)(a+b)=1 /(a+b), n=n$
Substitute the values in
$\begin{array}{l}
a\left(r^{n}-1\right) /(r-1)= \\
\quad=\left(a^{2}-b^{2}\right)\left(\frac{1-\left(\frac{1}{a+b}\right)^{n}}{1-\left(\frac{1}{a+b}\right)}\right) \\
=\left(a^{2}-b^{2}\right)\left(\frac{\left(\frac{(a+b)^{n}-1}{(a+b)^{n}}\right)}{\frac{(a+b)-1}{a+b}}\right) \\
=\frac{(a+b)(a-b)}{(a+b)^{n-1}}\left(\frac{(a+b)^{n}-1}{(a+b)-1}\right) \\
=\frac{(a-b)}{(a+b)^{n-2}}\left(\frac{(a+b)^{n}-1}{(a+b)-1}\right)
\end{array}$