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I'm asked to compute $$\sum_{k=-3}^{10} 2k^4$$ I looked up forBernoulli Number on Wikipedia and found a general formula for that. But my teacher has asked me to evaluate this by breaking the summation into the summations whose sum we already know i.e. say $\sum k^2$, arithmetic series, geometric series etc.

Can it be solved in this way, can anyone please tell me or give me some hint? It's really urgent and I can't do anything about it.

Edit : since there is a confusion, I want to clear it by saying that I'm not allowed to use the formula for $$\sum_{k=1}^n k^4$$. So I need an answer that doesn't include it.

Itachi
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  • Hint : $$\sum_{k=-3}^{10} 2k^4 = \sum_{n=1}^{14} 2(n-4)^4$$ – p_square Nov 26 '21 at 13:52
  • @Om3ga Yes I did that but expanding that still gives me the $\sum k^4$ term. Back to square one then. – Itachi Nov 26 '21 at 13:53
  • let me write an answer to explain it better – p_square Nov 26 '21 at 13:54
  • Thanks @Om3ga, but I'll suggest don't give me direct answer, help me solve it. – Itachi Nov 26 '21 at 13:55
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    ok will keep that in mind – p_square Nov 26 '21 at 13:55
  • posted the answer – p_square Nov 26 '21 at 13:59
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    Let $S_{m,n}=\sum_{k=1}^{n}k^m$. You can use this trick to compute $S_{4,n}=\sum_{k=1}^{n}k^4$, assuming that you already know the formulas for $S_{1,n},S_{2,n},S_{3,n}$. The same trick works to get these too. Start with $(n+1)^5=S_{5,n+1}-S_{5,n}=\sum_{k=1}^{n+1}k^5-\sum_{k=1}^{n}k^5=1+\sum_{k=1}^{n+1}(k+1)^5-\sum_{k=1}^{n}k^5=1+\sum_{k=1}^{n}[k^5+5k^4+10k^3+10k^2+5k+1-k^5]$. As you see the $k^5$ cancel and you get in the right hand side a combination of $S_{1,n}, S_{2,n},S_{3,n}$ and $S_{4,n}$. If you know the first three, you can solve for $S_{4,n}$. – plop Nov 26 '21 at 14:19
  • You've clarified you can't use the formula for $\sum k^4$ but can use the formula for $\sum k^2$. Which other formulas can you use? E.g., can you use the one for $\sum k^3$? – Barry Cipra Nov 26 '21 at 14:22
  • @Barry although I know the formula for $\sum k^3$ I'm only allowed till $\sum k^2$ – Itachi Nov 26 '21 at 14:30
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    @Itachi, you could, of course, simply do the calculation; there are only fourteen terms, after all, the largest of which is $2\cdot10^4=20{,}000$. Beyond that, I'm hard pressed to imagine any approach that doesn't effectively derive the formula for $\sum k^4$. – Barry Cipra Nov 26 '21 at 14:37

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$$\sum_{k=-3}^{10} 2k^4 = \sum_{n=1}^{14} 2(n-4)^4$$ now expand $2(n-4)^4$ which comes out to be : $$2n^4 - 32n^3 + 192n^2 - 512n + 512$$ Now our sum becomes : $$\sum_{n=1}^{14} 2n^4 - 32n^3 + 192n^2 - 512n + 512$$ Now after that apply the sum rule which says that $$\sum x_n + y_n = \sum x_n + \sum y_n$$ so our sum would become : $$\sum_{n=1}^{14} 2n^4 - \sum_{n=1}^{14} 32n^3 + \sum_{n=1}^{14} 192n^2 - \sum_{n=1}^{14} 512n + 512$$ Now I think you can take it from here

p_square
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    Actually I cant. This is what I was talking about earlier that in this expansion there is a term $\sum n^4$, but how can I compute it using the known results that I mentioned in my question? – Itachi Nov 26 '21 at 14:02
  • @Itachi didn't your teacher tell you about the formula for $\sum n^4$ ? – p_square Nov 26 '21 at 14:03
  • I can evaluate rest of the terms easily but what about this $\sum n^4$ part??? – Itachi Nov 26 '21 at 14:03
  • no actually he didn't. That's why the problem I'm facing. – Itachi Nov 26 '21 at 14:04
  • @Itachi the rule is that $$\sum_{k=1}^{n-1} n^4 =1/30(6n^5 + 15n^4 + 10n^3 - n)$$ – p_square Nov 26 '21 at 14:05
  • Please accept the answer if you can get to the final answer :) – p_square Nov 26 '21 at 14:06
  • thank you. But is not there any other way to compute that $\sum n^4$ other than this direct formula? And also can you provide me a link of the proof of this formula? – Itachi Nov 26 '21 at 14:07
  • I'll accept it don't worry. But first I need to get the answer I'm looking for. You must have understood what I'm talking about right? – Itachi Nov 26 '21 at 14:08
  • @Itachi this has it all – p_square Nov 26 '21 at 14:08
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    This just made it more complicated. You can compute $\sum_{i=1}^32k^4 + \sum_{i=0}^{10} 2k^4$ instead of breaking it into five different terms. – Kolja Nov 26 '21 at 14:09
  • @Itachi I think it is simpler to calculate $\sum_{k=0}^{10} 2k^4$ by using the formula and $\sum_{k=-3}^{-1} 2k^4$ by adding these three summands – miracle173 Nov 26 '21 at 14:10
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    I saw it @Om3ga. But what about my main question, isn't there any other way to evaluate $\sum n^4$ it without using this direct formula? I'm not allowed to do it actually. – Itachi Nov 26 '21 at 14:11
  • Yes @mircle but for that I'll need to use the formula for $\sum n^4$ but I'm not allowed to do that. Do you know any other way? – Itachi Nov 26 '21 at 14:13
  • @Itachi the best you can do is that derive it yourself or there could be another method which does not involve the $n^4$ term. – p_square Nov 26 '21 at 14:14
  • But like if the question itself contains $2k^4$ then how can he/her ask to solve it without using the formula for $n^4$ – p_square Nov 26 '21 at 14:16
  • I'm not sure, maybe he wants us to apply what we have learnt by far, that's why he didn't allow us. @Om3ga. – Itachi Nov 26 '21 at 14:17
  • @Itachi just a hint to derive it yourself : Use $$n^5 - (n-1)^5$$ – p_square Nov 26 '21 at 14:18
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    See https://math.stackexchange.com/questions/1949881/sum-of-fourth-powers-in-terms-of-sum-of-squares – Arthur Vause Nov 26 '21 at 14:19
  • Thanks @ArthurVause – p_square Nov 26 '21 at 14:20
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$\sum^n_{k=1} k^m$ is a linear function of the powers $\{n^m\}_{m=1}^{n+1}$. Therefore $$ \sum^n_{k=1}k^4=An+Bn^2+Cn^3+Dn^4+En^5 $$ We can plug in the numbers $n=1-5$ to get 5 equations: $$ 1=A+B+C+D+E\\ 17=2A+4B+8C+16D+32E\\ 98=3A+9B+27C+81D+243E\\ 354=4A+16B+64C+256D+1024E\\ 3479=5A+25B+125C+625D+3125E $$ which can be solved (with appropriate software) to get: $$ A=-\frac{1}{30}\\ B= 0\\ C=\frac{1}{3}\\ D=\frac{1}{2}\\ E=\frac{1}{5}\\ $$

am301
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Assume we have no electronic or mechanical device that assists us in doing the calculations and we are short of time and paper so that we want to avoid excessive calculations by pencil and paper and our mental arithmetic skills are rather limited than I would propose the following way to calculate the value:

$$(k-1)^4+k^4+(k+1)^4=3k^4+2 {4\choose 2}k^2+2=3k^4+12k^2+2=:f(k)$$ Then

$$\sum_{k=-3}^{10} k^4\\=2\sum_{k=1}^{3}k^4+\sum_{k=4}^{6}k^4+\sum_{k=7}^{9}k^4+10^4\\ =2f(2)+f(5)+f(8)+10000$$ For the following calculation note that $5^4=25^2$ can be easily calculated, because of the well known trick that to square a number with the decimal representation $(a5)_{10}$ we calculate $a(a+1)$ and append the digits $25$ on the resulting decimal number. So to calculate $25^2$ we calculate $2\cdot3=6$ and append $25$ to $6$ and get $625$. We also need $2^{12}$ which is also simple because $2^{12}=2^2\cdot 2^{10}=4\cdot1024$ and $1024$ should be be known at least if one has attended a course in computer science.
$$ =2(3\cdot16+12\cdot4+2)+(3\cdot625+12\cdot25+2)+(3\cdot4096+12\cdot64+2)\\ =3(32+625+4096)+12(8+25+64)+8+10000\\ =(4653\cdot3+12\cdot 97+8+10000)\\ =13959+12(100-3)+10008\\ =23967+1200-36\\ =25131$$

To get the requested number we have to multiply this by $2$, but this is left to the reader.

miracle173
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