Sequences and S...

Find the sum of the following geometric series:
$$ 1,-a,a^2,-a^3,...$$ to n terms $$\left ( a\neq 1 \right )$$

In a sequence, $$a_{n}=n^{2}-1$$ then $$a_{n+1} $$ is equal to

Sum of the series 
$$S=1+\dfrac{1}{2} \left ( 1+2 \right )+\dfrac{1}{3}\left ( 1+2+3 \right )+\dfrac{1}{4}\left ( 1+2+3+4 \right )+...$$ upto $$20$$ terms is

The sum to infinity of $$\displaystyle\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+...$$is

$$3.6+6.9+9.12+...+3\mathrm{n}(3\mathrm{n}+3)=$$

$$2\cdot1^{2}+3\cdot2^{2}+4\cdot3^{2}+\dots $$ up to $$n$$ terms $$=$$

$$1.4+2.5+...+\mathrm{n}(\mathrm{n}+3)=$$

$$ 1+ 3 + 6 + 10 + ...+\displaystyle \frac{(n-1)n}{2}+\frac{n(n+1)}{2}=$$

If $$S_1,S_2$$ and $$S_3$$. are the sums of first n natural  numbers, their squares and their cubes respectively, then $$S_3\left (1+8S_1  \right )=$$

$$ 1.3+3.5+5.7+...+(2\mathrm{n}-1)(2\mathrm{n}+1)=$$