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trying to solve $\sum_{i=3}^{n+1} i$

First I attempt to change the lower/upper bounds:

$\sum_{i=3}^{n+1} i = \sum_{i=1}^{n-1} i$

in order to use $\sum_{i=1}^{n} i = \frac{n(n+1)}2$

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

This is just practice out of a textbook that doesn't have answers - but I tried to input the summation in wolframalpha and my result is not one of the answers there. Where have I messed up?

Additionally, is modifying the lower/upper bound of a summation in order to use an equality like the one above an ok way to approach these problems?

  • 0
    You made a mistake when changing the bounds, as you did not change the value of $i$ itself. Think of it like this: The first element to the left is $3$. After the change of order what do I need to change at the right side so that the first element will still be $3$? (And what is it now?) – 2017-01-12

3 Answers 3

4

You are probably making things complicate. It is nothing but $$ \biggr(\sum_{k=1}^nk\biggr)-1-2+(n+1) $$ and you have already known the formula for $\sum_{k=1}^nk$.

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    Interesting that accepted answer shows how to use formula but does not answer any of the real questions 1. **Where have I messed up** 2. **[how to modify] the lower/upper bound of a summation**. Obviously if formula for upper bound $n$ is known then one for $n+1$ is known also. – 2017-01-12
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    One might say the "real question" is in the **title**: Solving $\sum_{i=3}^{n+1}i$ and OP is thinking about the problem in a more complicate than necessary way. – 2017-01-12
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    Title has to be concise and is only indication of what will be in the question. Also using formula for upper bound $n$ only is making things more complicate than necessary. OP has realized that he can use sum of first $n$ numbers even if $n$ is $n+1$ or $n+999$ : just replace in the formula. Will you solve $\displaystyle\sum_{i=76}^{n+999}i$ by searching to get $\displaystyle\sum_{i=i}^{n}i$ ? Splitting $$\sum_{i=76}^{n+999} i = \sum_{i=1}^{n+999} i-\sum_{i=1}^{75} i$$ is solved readily by two applications of the formula. – 2017-01-12
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    Also writing $$\sum_{i=76}^{n+999} i=\sum_{m=1}^{n+924} (m+75)$$ requires only one application of the formula and sticks with original idea of OP. Then $$ \sum_{m=1}^{n+924} (m+75)=\sum_{m=1}^{n+924} m+\sum_{m=1}^{n+924} 75=\frac{(n+924)(n+925)}{2}+(n+924)\cdot75 $$ – 2017-01-12
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That's wrong. Take $n=2$. Then you say $3=1$.

Use $$ \sum_{i=3}^{n+1} i = \sum_{i=1}^{n+1} i-\sum_{i=1}^{2} i $$

If you want change order of summation, then set $m=i-2$. Then $m$ goes from $1$ to $n-1$ but $i=m+2$ so $$ \sum_{i=3}^{n+1} i=\sum_{m=1}^{n-1} (m+2) $$

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When you subtract 2 from the lower and upper bounds, you have to replace $i$ with $(i+2)$ wherever it appears in the summands to compensate.

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

Can you take things from here?