Calculus

Series

Frequently used series:

$$\sqrt{1+x}$$	0 ≤ x ≤ 1	$$= \sum _{ {n=0} }^{\infty }{\frac {(-1)^{n}(2n)!}{(1-2n)(n!)^{2}(4^{n})}}x^{n}$$	$$= 1+\textstyle {\frac {1}{2}}x-{\frac {1}{8}}x^{2}+{\frac {1}{16}}x^{3}-{\frac {5}{128}}x^{4}+\dots ,\!$$	[1]
$$\frac{1}{1+x}$$	-1 < x < 1	$$= \sum _{ {n=0} }^{\infty }x^{n}$$	$$= 1+x+x^{2}+x^{3}+\cdots \!$$

$$\sum_{i=1}^n i^2$$	$$\frac{1}{6}n(2n-1)(n-1)$$	See [2]. Basically, develop $\sum_{i=1}^n (i^3+3i^2+3i+1)=\sum_{i=1}^n (i+1)^3=\sum_{i=2}^{n+1} i^3=\sum_{i=1}^n i^3+(n+1)^3-1$, then isolate and simplify