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


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?

  • $\begingroup$ mathworld.wolfram.com/PowerSum.html $\endgroup$ – Jane Aug 25 '16 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 Aug 25 '16 at 17:08
  • $\begingroup$ @Jane You were a bit faster :) $\endgroup$ – Peter Aug 25 '16 at 17:09
  • 1
    $\begingroup$ @Peter Multiplication dots $\endgroup$ – MathGeek Aug 25 '16 at 17:11
  • 1
    $\begingroup$ 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 Aug 25 '16 at 17:12

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: $$ \sum_{n=1}^{N}n(n+1)^2(n+2) = \color{red}{\frac{1}{10} N (1+N) (2+N) (3+N) (3+2 N)}.\tag{3}$$


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,

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

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



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

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

Adding column wise we get,


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

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:

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


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.


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 $




  • 7
    $\begingroup$ I have trouble understanding why you would call this a "hint"... $\endgroup$ – Arthur Aug 25 '16 at 17:19
  • 1
    $\begingroup$ Because it is up to the OP to prove this statement. $\endgroup$ – Jan Aug 25 '16 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.