$$\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) $$
I thought I would use the distributive rule but the question stipulates that my answer should not include a summation sign...
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For every term the two sums have in common, you get a $3$. Then there are two terms left over. This is why you don't need a summation sign – Matt Samuel Sep 14 '16 at 00:05
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Method 1:
$$\begin{align} \sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & =\sum_{i=1}^n (3i^2 +4) - \sum_{j=1}^n (3(j+1)^2 +1) \\ & = \sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2+6j+4) \\ & = \sum_{i=1}^n \left[(3i^2 +4) - (3i^2+6i+4)\right] \\ & = \sum_{i=1}^n -6i \\ & = -6\cdot\frac{n(n+1)}2 \\ \sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & =-3n(n+1) \\ \end{align}$$
Method 2:
$$\begin{align} \sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & = 7+\sum_{i=2}^n (3i^2 +4) - \sum_{j=2}^n (3j^2 +1)-3(n+1)^2-1 \\ & =6-3(n+1)^2+\sum_{i=2}^n \left[(3i^2 +4) - (3i^2 +1)\right] \\ & =6-3(n+1)^2+\sum_{i=2}^n3 \\ & =6-3(n+1)^2+\underbrace{3+3+3+\dots+3}_{n-1} \\ & =6-3(n+1)^2+3(n-1) \\ & =-3n^2-3n \\ \sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) & =-3n(n+1) \\ \end{align}$$
Simply Beautiful Art
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Thanks. Could you explain conceptually the leap from summation to simple equation. I think I understand: Since i is just a place holder and the first sum is n and the second n+1 and it becomes all over 2 since it's subtraction, then why do you multiple n & n+1. – SunshineTS Sep 14 '16 at 00:08
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How so? I'm really trying to understand these conceptually and not just mathematically. – SunshineTS Sep 14 '16 at 00:09
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@SunshineTS It is the result of $\sum_{i=1}^ni=\frac{n(n+1)}2$, which has interesting proofs. – Simply Beautiful Art Sep 14 '16 at 00:09
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1@SunshineTS What do you mean "conceptually"? I'll happily expand my answer based on what you want. – Simply Beautiful Art Sep 14 '16 at 00:10
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Appreciate it. $\sum_{i=1}^ni=\frac{n(n+1)}2$ Where does this come from? I think I am just confusnig sums and sets in my head. – SunshineTS Sep 14 '16 at 00:18
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1@SunshineTS I updated the answer to include MattSamuel's insights, and $\sum_{i=1}^ni=\frac{n(n+1)}2$ proof can be found here: http://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-ldotsn – Simply Beautiful Art Sep 14 '16 at 00:25
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