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Find the sum of $n$ terms of following series:

$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$

I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way find the sum?

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  • $\begingroup$ mathworld.wolfram.com/PowerSum.html $\endgroup$
    – Kerr
    Commented Aug 25, 2016 at 17:07
  • $\begingroup$ The sum of the $n^4$ is not really problem : wolframalpha.com/input/?i=sum(k%3D1,n,k%5E4) $\endgroup$
    – Peter
    Commented Aug 25, 2016 at 17:08
  • $\begingroup$ @Jane You were a bit faster :) $\endgroup$
    – Peter
    Commented Aug 25, 2016 at 17:09
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    $\begingroup$ @Peter Multiplication dots $\endgroup$
    – MathGeek
    Commented Aug 25, 2016 at 17:11
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    $\begingroup$ Anyway, I suggest to compute it using $$\sum_{k=2}^{n+1} (k-1)(k+1)k^2 = \sum_{k=1}^{n+1} k^4 - k^2 = \sum_{k=1}^{n+1} k^4 - \sum_{k=1}^{n+1}k^2$$ $\endgroup$
    – Crostul
    Commented Aug 25, 2016 at 17:12

4 Answers 4

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We have $$ n(n+1)^2(n+2) = 24\binom{n+2}{4}+12\binom{n+2}{3}\tag{1}$$ hence $$ \sum_{n=1}^{N}n(n+1)^2(n+2) = 24\binom{N+3}{5}+12\binom{N+3}{4} \tag{2}$$ is a consequence of the Hockey Stick identity, leading to: $$ \sum_{n=1}^{N}n(n+1)^2(n+2) = \color{red}{\frac{1}{10} N (1+N) (2+N) (3+N) (3+2 N)}.\tag{3}$$

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To find $\sum_{k=1}^{n}(k^4)$ you may follow this process.

Consider the identity, $(x+1)^5-x^5=5x^4+10x^3+10x^2+5x+1$.

Putting $x=1,2,3,...,(n-1),n$ successively, we get,

$2^5\space-\space 1^5=5\cdot1^4+10 \cdot1^3+10\cdot1^2+5\cdot1+1$

$3^5\space-\space 2^5=5\cdot2^4+10 \cdot2^3+10\cdot2^2+5\cdot2+1$

$\cdot\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\cdot$

$\cdot\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\cdot$

$n^5\space-\space (n-1)^5=5\cdot(n-1)^4+10 \cdot(n-1)^3+10\cdot(n-1)^2+5\cdot(n-1)+1$

$(n+1)^5\space-\space n^5=5\cdot n^4+10 \cdot n^3+10\cdot n^2+5\cdot n+1$

Adding column wise we get,

$(n+1)^5-1^5=5(1^4+2^4+\cdot\cdot\cdot+n^4)+10(1^3+2^3+\cdot\cdot\cdot+n^3)+10(1^2+2^2+\cdot\cdot\cdot+n^2)+5(1+2+\cdot\cdot\cdot+n)+(1+1+\cdot\cdot\cdot+n)$

$\implies n^5+5n^4+10n^3+10n^2+5n=5\sum_{k=1}^{n}(k^4)+10\sum_{k=1}^{n}(k^3)+10\sum_{k=1}^{n}(k^2)+5\sum_{k=1}^{n}(k)+n$

Knowing the sum of the series' $\sum n^3$ , $\sum n^2$ and $\sum n$, you can solve for $\sum n^4$ from the above equation to get the result:

$\sum n^4=\frac {n(n+1)(2n+1)(3n^2+3n-1)}{30}$

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An alternative to finding the formula for $\sum k^4$ is to instead write the $k$th term as $$(k+2)(k+1)^2k=(k+3)(k+2)(k+1) k-2(k+2)(k+1)k.$$ That is, one trades sums of integer powers for sums of consecutive products. To see the advantage, note for example that

$$1\cdot 2\cdot 3 +2\cdot 3\cdot 4+3\cdot 4\cdot 5+4\cdot 5\cdot 6=210= \frac{1}{4}(4\cdot 5\cdot 6\cdot 7),$$ i.e. the sum of products of three consecutive integers can be expressed in terms of a product of four consecutive integers. A similar pattern holds for other such sums, allowing the required formulas to be readily found. (By comparison, the pattern for sums of integer powers is much more involved.) From such computations one can then obtain the desired formula for the original sum.

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The answer is $S_n = \frac{1}{5}n^5 + \frac{3}{2}n^4 + 4n^3 + \frac{9}{2}n^2 + \frac{54}{30}n $

$\sum n^4 $ is not really a problem.

You can do it using this method here https://www.coursera.org/learn/calculus1/lecture/uR7YK/what-is-the-sum-of-n-4-for-n-1-to-n-k

I don't know if this is helpful but this is how i got it

$S_n = \sum a_n$

$a_n = n(2+n)(1+n)^2 = n^4+4n^3+5n^2+2n$

therefore : $ S_n = \sum {(n^4+4n^3+5n^2+2n)} $

$S_n = \sum n^4 + 4\sum n^3 + 5\sum n^2 + 2\sum n $

$S_n = \frac{1}{30}( 6n^5 +15n^4 + 10n^3- n) +4(\frac{1}{4}(n^4+2n^3+n^2) + 5(\frac{1}{6}(2n^3 + 3n^2 +n ) ) + 2(\frac{1}{2}(n^2 +n ) ) $

hence, $S_n = \frac{1}{5}n^5 + \frac{3}{2}n^4 + 4n^3 + \frac{9}{2}n^2 + \frac{54}{30}n $

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