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I know this is something that will pop up for my math exam tomorrow. When searching for this the only thing I could find was this: Calculate $\sum\limits_{n=1}^\infty (n-1)/10^n$ using pen and paper, but this is for infinite sums. So I'd like to find something about how to calculate a finite sum, typically from 1 to 100. This is something a computer could easily calculate in Maple or MATLAB etc. but we are not allowed to using math programs, not even a calculator, pen and paper only!

Examples from old exams are: Calculate the sum: $$\sum_{i=1}^{100}(2i+1)$$

Another example: Calculate the sum: $$\sum_{i=1}^{99}(6i^2+2i)$$

The only way to do it, that I can think of would be like this:

$$\sum_{i=1}^{100}(2i+1)=(2\cdot1+1)+(2\cdot2+1)+(2\cdot3+1)+(2\cdot4+1)+\dots+(2\cdot100+1)=10200$$

However, this would take way too long. I am sure there are better ways to solve these kinds of problem, but how?

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  • $\begingroup$ take a look here $\endgroup$ – Masacroso May 28 '17 at 11:05
  • $\begingroup$ Most people seem to prefer stackexchange to pen and paper. $\endgroup$ – Professor Vector May 28 '17 at 11:12
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For the first use the triangular number formula: $$\sum_{i=1}^{100}(2i+1) = 2\sum_{i=1}^{100}i+\sum_{i=1}^{100}1 $$$$= 2\frac{100(100+1)}{2}+100 = 10100 + 100 $$$$= 10200$$

For the second use the formula for square pyramidal numbers as well:

$$\sum_{i=1}^{99}(6i^2+2i) = 6\sum_{i=1}^{99}i^2+2\sum_{i=1}^{99}i $$ $$= 6\frac 1 6 99(99+1)(198+1) + 2\frac{99(99+1)}{2} $$ $$= (198+1+1)99(99+1) = 200\cdot100\cdot 99 $$ $$ = 1980000$$

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hint

For $|x|<1$,

$$\sum_0^\infty x^n=1/(1-x) $$

by differentiation $$\sum_1^\infty nx^{n-1}=1/(1-x)^2$$

multilying by $x $

$$\sum_1^\infty nx^n=x/(1-x)^2$$

replace $x=1/10$.

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The first one is just an Arithmetic Progression!

Use $\frac n2 (a+\ell)$.

$$\sum_{i=1}^{100}(2i+1)=\frac {100}2 \big(3+201\big)=\color{red}{10200}$$


For the second one, the coefficients give a useful clue to a shortcut.

Note that $\displaystyle\sum_{i=1}^n i^2=\frac {n(n+1)(2n+1)}6$ and $\displaystyle\sum_{i=1}^n i=\frac {n(n+1)}2$.

Hence

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

Putting $n=99$ gives $$\sum_{i=1}^{99}=2\cdot 99\cdot 100^2=\color{red}{1980000}$$

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