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I want to simplify

$$\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)$$

without a summation sign in the answer. How can I do this?

I can't work out how to simplify this expression. I appreciate any help.

Caleb Stanford
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John434
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  • Use this formula: http://math.stackexchange.com/questions/48080/prove-that-sum-limits-k-1nk2-fracnn12n16?noredirect=1&lq=1 – Alexis Olson Sep 13 '16 at 05:22
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    Have you tried with $n = 3$ or $4$, and seen what happens? @AlexisOlson That's unnecessary, really. The answer is much simpler than that. – Arthur Sep 13 '16 at 05:22
  • Oh wow. It's a minus sign. That does make things simpler. – Alexis Olson Sep 13 '16 at 05:23
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    Hint: you can distribute the summation so that you have two sums of constants and two sums with variable dependence (of the form $\Sigma 3i^2$). Notice that a lot of the terms with variable dependence cancel – JMK Sep 13 '16 at 05:24

4 Answers4

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Well the obvious first step will to try to see if we can combine terms:

$\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)=$

$(3\cdot1^2 + 4) + \sum_{k=2}^n (3k^2+4) - \sum_{k=2}^n (3k^2+1) - [3(n+1)^2 + 1]=$

$(3\cdot1^2 + 4)+\sum_{k=2}^n[(3k^2+4)- (3k^2+1)] - [3(n+1)^2 + 1]$

and see what happens. As it turns out a lot happens. (All of it good.)

Arthur
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fleablood
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Hint: Sometimes it helps to write the terms of a sum so you're not just looking at symbols. The sum is \begin{align*} (3 \cdot 1^2 + 4) + &(3 \cdot 2^2 + 4) + (3 \cdot 3^2 + 4) + \cdots + (3 \cdot n^2 + 4) \\ - \Big[ &(3 \cdot 2^2 + 1) + (3 \cdot 3^2 + 1) + \cdots + (3 \cdot n^2 + 1) + (3 \cdot (n+1)^2 + 1)\Big]. \end{align*}

Can you see how to simplify that?

Caleb Stanford
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That solves it. Expand the terms ..... enter image description here

tantheta
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$$\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)=\sum_{i=1}^n (3i^2+4) - \sum_{j=1}^{n} (3(j+1)^2+1)$$ $$=\sum_{i=1}^n (3i^2+4) - \sum_{i=1}^{n} (3(i+1)^2+1)=\sum_{i=1}^n (3i^2+4) - \sum_{i=1}^{n} 3i^2+6i+4$$ $$=\sum_{i=1}^{n} -6i=-6\frac{n}{2}(n+1)=-3n(n+1)$$

E.H.E
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