# Rules for Product and Summation Notation

When we deal with summation notation, there are some useful computational shortcuts, e.g.: $$\sum\limits_{i=1}^{n} (2 + 3i) = \sum\limits_{i=1}^{n} 2 + \sum\limits_{i=1}^{n} 3i = 2n + \sum\limits_{i=1}^{n}3i$$

However, I don't think I know all the useful shortcuts here. Are there other computational tricks one should be aware of? What's a good way for thinking about this?

More importantly, consider product notation: $$\prod\limits_{i=1}^{n} (\sqrt{2} - \sqrt[n]{2})$$

I don't know what the shortcuts here are. What are some of the more effective ways of attacking such computations?

Some sum identities:

$$\sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n)$$

$$\sum_{n=s}^t f(n) + \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) + g(n)\right]$$

$$\sum_{n=s}^t f(n) - \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) - g(n)\right]$$

$$\sum_{n=s}^t f(n) = \sum_{n=s+p}^{t+p} f(n-p)$$

$$\sum_{n=s}^j f(n) + \sum_{n=j+1}^t f(n) = \sum_{n=s}^t f(n)$$

$$\sum_{n\in A} f(n) = \sum_{n\in \sigma(A)} f(n)$$

$$\sum_{i=k_0}^{k_1}\sum_{j=l_0}^{l_1} a_{i,j} = \sum_{j=l_0}^{l_1}\sum_{i=k_0}^{k_1} a_{i,j}$$

$$\sum_{n=0}^t f(2n) + \sum_{n=0}^t f(2n+1) = \sum_{n=0}^{2t+1} f(n)$$

$$\sum_{n=0}^t \sum_{i=0}^{z-1} f(z\cdot n+i) = \sum_{n=0}^{z\cdot t+z-1} f(n)$$

$$\sum_{n=s}^t \ln f(n) = \ln \prod_{n=s}^t f(n)$$

$$c^{\left[\sum_{n=s}^t f(n) \right]} = \prod_{n=s}^t c^{f(n)}$$

$$\sum_{i=m}^n 1 = n+1-m$$

$$\sum_{i=1}^n \frac{1}{i} = H_n$$

$$\sum_{i=1}^n \frac{1}{i^k} = H^k_n$$

$$\sum_{i=m}^n i = \frac{n(n+1)}{2} - \frac{m(m-1)}{2} = \frac{(n+1-m)(n+m)}{2}$$

$$\sum_{i=0}^n i = \sum_{i=1}^n i = \frac{n(n+1)}{2}$$

$$\sum_{i=0}^n i^2 = \frac{n(n+1)(2n+1)}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$$

$$\sum_{i=0}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4} = \left[\sum_{i=1}^n i\right]^2$$

$$\sum_{i=0}^n i^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30}$$

$$\sum_{i=0}^n i^p = \frac{(n+1)^{p+1}}{p+1} + \sum_{k=1}^p\frac{B_k}{p-k+1}{p\choose k}(n+1)^{p-k+1}$$

$$\left(\sum_{i=m}^n i\right)^2 = \sum_{i=m}^n ( i^3 - im(m-1) )$$

$$\sum_{i=m}^n i^3 = \left(\sum_{i=m}^n i\right)^2 + m(m-1)\sum_{i=m}^n i$$

$$\sum_{i=m}^{n-1} a^i = \frac{a^m-a^n}{1-a}$$

$$\sum_{i=0}^{n-1} a^i = \frac{1-a^n}{1-a}$$

$$\sum_{i=0}^{n-1} i a^i = \frac{a-na^n+(n-1)a^{n+1}}{(1-a)^2}$$

$$\sum_{i=0}^{n-1} i 2^i = 2+(n-2)2^{n}$$

$$\sum_{i=0}^{n-1} \frac{i}{2^i} = 2-\frac{n+1}{2^{n-1}}$$

$$\sum_{i=0}^n {n \choose i} = 2^n$$

$$\sum_{i=1}^{n} i{n \choose i} = n2^{n-1}$$

$$\sum_{i=0}^{n} i!\cdot{n \choose i} = \sum_{i=0}^{n} {}_{n}P_{i} = \lfloor n!\cdot e \rfloor$$

$$\sum_{i=0}^{n-1} {i \choose k} = {n \choose k+1}$$

$$\sum_{i=0}^n {n \choose i}a^{(n-i)} b^i=(a + b)^n$$

$$\sum_{i=0}^n i\cdot i! = (n+1)! - 1$$

$$\sum_{i=1}^n {}_{i+k}P_{k+1} = \sum_{i=1}^n \prod_{j=0}^k (i+j) = \frac{(n+k+1)!}{(n-1)!(k+2)}$$

$$\sum_{i=0}^n {m+i-1 \choose i} = {m+n \choose n}$$

• Oh wow. That's very thorough. Thank You. – Newb Nov 4 '13 at 10:48
• Did you copy those from somewhere? What about product identities? – endolith Dec 20 '13 at 3:04
• Nah - it's obvious that LTS just entered them in the spur of the moment off the top of her/his head. However, I suggest you save these, because they will prove very useful. – marty cohen Dec 11 '14 at 6:21

Summation of product of two functions.

$$\sum_{i=1}^{n} f(x) g(x)$$ = $$\sum_{i=1}^{n}f(x)∑g(x) - [ f(x)∑g(x-1) + f(x-1)∑g(x-2) + f(x-2)∑g(x-3) + .... + f(2)∑g(1) + g(x)∑f(x-1) + g(x-1)∑f(x-2) + g(x-2)∑f(x-3) + .... + g(2)∑f(1)]$$

Here are the formula for the sum of the first $n$ natural numbers and the first $n$ squares. There are similar formula for the sum of the first $n$ cubes etc...
$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$ $$\sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}$$
if we know the first term $a_1$ and the last term $a_n$, and the series has $n$ terms, then the sum will be $$\frac{n(a_1+a_n)}{2}$$
I suppose $$\prod\limits_{i=1}^{n}(x)$$ means multiplying from 1 to n, which is n factorial. So maybe it will help. Also $$(\sqrt{2} - \sqrt[n]{2})$$ does not contain any i.