I want a 'simple' proof to show that:  $$1^4+2^4+...+n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$  I tried to prove it like the others but I can't and now I really need the proof. Also I want a geometric proof for that and this one below:  $$1^3+2^3+3^3+...=(\frac{n(n+1)}{2})^2$$  geometric proof: I want a proof using shapes and geometry.
120478
Summing the first $n$ first powers of natural numbers: $$\sum_{k=1}^nk=\frac12n(n+1)$$ and there is a  involving two copies of a 2D representation of $(1+2+\cdots+n)$ that form a $n\times(n+1)$-rectangle.    Similarly, $$\sum_{k=1}^nk^2=\frac16n(n+1)(2n+1)$$  has a  (scroll down just a bit til you see the wooden blocks) involving six copies of a 3D representation of $(1^2+2^2+\cdots+n^2)$ that form a $n\times(n+1)\times(2n+1)$-rectangular solid.     And similarly, $$\sum_{k=1}^nk^3=\frac14n^2(n+1)^2$$ has a proof involving four copies of a 4D representation of $(1^3+2^3+\cdots+n^3)$ that form a $n\times n\times(n+1)\times(n+1)$-rectangular hypersolid in four-space. I can't find a resource to demonstrate this last one better, but I've sketched it out for $n=3$, sketching the four dimensions using a $4\times4$ grid for two dimensions, and within each cell a $3\times 3$ grid for the other two. (Try it - it's fun!) Here's a smaller $n=2$ version. Note that $1^3+2^3$ is represented by some 4D blocks $1\times1^3+1\times2^3$ that are configured in a way that makes use of all four dimensions. $$\begin{array}{c|c|c} \begin{array}{cc} {\color{red} \bullet} &amp; {\color{red} \bullet}\\ {\color{red} \bullet} &amp; {\color{red} \bullet} \end{array} &amp; \begin{array}{cc} {\color{red} \bullet} &amp; {\color{red} \bullet}\\ {\color{red} \bullet} &amp; {\color{red} \bullet} \end{array} &amp; \begin{array}{cc} {\color{blue} \bullet} &amp; {\color{blue} \bullet}\\ {\color{blue} \bullet} &amp; {\color{blue} \bullet} \end{array} \\\hline \begin{array}{cc} {\color{green} \bullet} &amp; {\color{green} \bullet}\\ {\color{green} \bullet} &amp; {\color{green} \bullet} \end{array} &amp; \begin{array}{cc} {\color{red} \bullet} &amp; {\color{blue} \bullet}\\ {\color{green} \bullet} &amp; {\color{magenta} \bullet} \end{array} &amp; \begin{array}{cc} {\color{blue} \bullet} &amp; {\color{blue} \bullet}\\ {\color{blue} \bullet} &amp; {\color{blue} \bullet} \end{array} \\\hline \begin{array}{cc} {\color{green} \bullet} &amp; {\color{green} \bullet}\\ {\color{green} \bullet} &amp; {\color{green} \bullet} \end{array} &amp; \begin{array}{cc} {\color{magenta} \bullet} &amp; {\color{magenta} \bullet}\\ {\color{magenta} \bullet} &amp; {\color{magenta} \bullet} \end{array} &amp; \begin{array}{cc} {\color{magenta} \bullet} &amp; {\color{magenta} \bullet}\\ {\color{magenta} \bullet} &amp; {\color{magenta} \bullet} \end{array} \\ \end{array}$$ This is a $2\times2\times3\times3$ rectangular hypersolid made up of four copies of $1\cdot1^3+1\cdot2^3$.    So we move on to fourth powers: $$\sum_{k=1}^nk^4=\frac1{30}n(n+1)(2n+1)(3n^2+3n-1)$$ This time the polynomial does not completely factor. I would like to know if anyone can find a similar geometric proof. It would involve 30 copies of a 5D representation of $(1^4+2^4+\cdots+n^4)$ forming a 5D hypersolid, one of whose 2-faces somehow works out to have area $3n^2+3n-1$, with the orthogonal edge lengths being $n$, $n+1$, and $2n+1$. I know it seems hopeless...30 copies? $3n^2+3n-1$? But it would make for some nice art.  It would be a nice start if there were some sort of connection between $30$ and a symmetry group of $\mathbb{R}^5$. For instance, if there were a subgroup of $\operatorname{SO}_5$ whose order was some large divisor of $30$, then that might help place the $30$ blocks of volume $1^5$ into their configuration.
49962
I need to prove that $$ 2^n &gt; n^3\quad \forall n\in \mathbb N, \;n&gt;9.$$  Now that is actually very easy if we prove it for real numbers using calculus. But I need a proof that uses mathematical induction.   I tried the problem for a long time, but got stuck at one step - I have to prove that:     $$ k^3 &gt; 3k^2 + 3k + 1 $$   Hints???
68551
Use mathematical induction to show that:  $1^3 + 3^3 + 5^3 + ... + (2n + 1)^3 = (n+1)^2(2n^2+4n+1)$ whenever $n$ is a positive integer.  I am able to solve this problem, but the thing that confuses me on this is the Basis Step.  P(1): $(2(1) + 1)^3$ $\neq$ $4(7)$, does this mean that this problem can not be proved using mathematical induction, or am I missing something? Thanks.
102016
As the title says I need to prove the following by induction: $$\sum_{i=1}^{2^n} \frac{1}{i} \ge 1+\frac{n}{2}, \forall n \in \mathbb N$$   When trying to prove that P(n+1) is true if P(n) is, then I get that:  $$\sum_{i=1}^{2^{n+1}} \frac{1}{i} = (\sum_{i=1}^{2^n} \frac{1}{i} ) +(\sum_{i=1}^{2^n} \frac{1}{2^n+i} ) \ge 1+\frac{n}{2} + (\sum_{i=1}^{2^n} \frac{1}{2^n+i} )$$ using the inductive hypothesis. And for P(n+1) to be true it would be sufficient to prove that  $$1+\frac{n}{2} + (\sum_{i=1}^{2^n} \frac{1}{2^n+i} ) \ge 1+\frac{n+1}{2}$$  or rewriting $$\sum_{i=1}^{2^n} \frac{1}{2^n+i} \ge \frac{1}{2} $$ and here I'm a bit stuck  Thank you for your help!
18200
How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}&lt;\binom{2n}{n}&lt;\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ &gt; 1 ?     Does anyone know any better elementary estimates?   Attempt.  We have $$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}=\prod_{k=0}^{n-1}\left(1+\frac{k}{2(n-k)}\right).$$ Then we have $$\left(1+\frac{k}{2(n-k)}\right)&gt;\sqrt{1+\frac{k}{n-k}}=\frac{\sqrt{n}}{\sqrt{n-k}}.$$ So maybe, for the lower bound, we have $$\frac{n^{\frac{n}{2}}}{\sqrt{n!}}=\prod_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n-k}}&gt;\frac{2^n}{\sqrt{4n}}.$$ By Stirling, $n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, so the lhs becomes $$\frac{e^{\frac{n}{2}}}{(2\pi n)^{\frac14}},$$ but this isn't $&gt;\frac{2^n}{\sqrt{4n}}$.
50237
I know that $$1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2$$  What I would like to know is whether there is a simple proof (that obviously does not use the above info) as to why the sum of the first $n$ cubes always a perfect square.
115203
Prove that for $n \in \mathbb{N}$  (a) $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\geq \frac{2}{3}$  And   (b) $\frac{1}{2} \leq \frac{1}{3n+1}+\frac{1}{3n+2}+.....\frac{1}{5n}+\frac{1}{5n+1}+....&lt; \frac{2}{3}$  my attempt :  $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\\ &gt;\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2n}+....+\frac{1}{2n}\\ =\frac{n+1}{2n}&gt;\frac{1}{2}$  Where i am missing and how to processed second question
15333
How to prove that $ 2&lt;(1+\frac{1}{n})^{n}$ for every integer $n&gt;1$. I was thinking by induction, it works for $n=2$ but then I couldn't move forward.
65526
Are there any positive $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square?  I tried to simplify   \begin{align*} n^4+n^3+n^2+n+1 &amp;= n^2(n^2+1)+n(n^2+1)+1\\  &amp;= (n^2+n)(n^2+1)+1 \\   &amp;= n(n+1)(n^2+1)+1 \end{align*} Then I assumed that the above expression is a square; then  $$ n(n+1)(n^2+1)+1 = k^2$$  $$  \begin{align*} n(n+1)(n^2+1) &amp;= (k^2-1) \\    &amp;= (k+1)(k-1) \end{align*} $$   Then trying to reason with prime factors, but cannot find a concrete proof yet.
94918
I've gotten to the final step and believe my problem lies within my algebra.  Prove the following:   $1 \times 3 + 2 \times 4 + 3 \times 5 + ... + N(N+2) = \frac{N(N+1)(2N+7)}6$  Here is my show that it is true answer:   $\displaystyle\frac{(K+1)[(k+1)+1][(2(k+1)+7]}6$  Here is where I am stuck:  K(K+1)(2K+7) + 6 (k+1) [(k+1)+2]/6  So all that needs to be done is algebra and moving around the above to show that it is equal to the show that it is true answer above. I just cannot figure it out. I am pretty sure I have the right equation.
48695
I'm having some trouble figuring out where to start for this proof.     Prove for all positive integers $n$,   $$\sqrt{n+1}-\sqrt{n}\lt \frac{1}{2\sqrt{n}}$$   Any help will be greatly appreciated.
115709
Proving by induction. We'd like to show that $2 + 4 + 6 + \cdots+ 2n = n(n + 1)$.  A nice way to do this is by induction. Let $S(n)$ be the statement above. An inductive proof would have the following steps:  Show that $S(1)$ is true.  Show that if $S(1),\ldots,S(k)$ are true, then so is $S(k + 1)$.    This question is really starting to bug me, am I doing something wrong or is the equation wrong for that series? It only seems to work for $S(1)$ but after that it does not give the correct series. The $n(n+1)$ series is $2, 6, 12,\ldots.$ I'm really confused can someone please nudge me in the right direction (I know how proofs by induction works) I'm just having problems with this one in particular.
110441
I know how to prove this by induction but the text I'm following shows another way to prove it and I guess this way is used again in the future. I'm confused by it.  So the expression for first n numbers is: $$\frac{n(n+1)}{2}$$  And this second proof starts out like this. It says since:  $$(n+1)^2-n^2=2n+1$$  Absolutely no idea where this expression came from, doesn't explain where it came from either.  Then it proceeds to say: \begin{align} 2^2-1^2&amp;=2*1+1 \\ 3^2-2^2&amp;=2*2+1\\ &amp;\dots\\ n^2-(n-1)^2&amp;=2(n-1)+1\\ (n+1)^2-n^2&amp;=2n+1 \end{align} At this point I'm completely lost. But it continues to say "adding and noting the cancellations on the left, we get" \begin{align} (n+1)^2-1&amp;=2(1+2+...+n)+n \\ n^2+n&amp;=2(1+2+...+n) \\ (n(n+1))/2&amp;=1+2+...+n \end{align}  Which proves it but I have no clue what has happened. I am entirely new to these math proofs. Im completely lost. I was great at high school math and calculus but now I haven't got the slightest clue of what's going on. Thanks
21280
I just want to know if my proof to this question is correct. First, I proved it was true for $n = 4$. $$4!&gt;2^4$$ $$24&gt;16$$ Then, I assumed that it was true for $n=k$. $$k!&gt;2^k$$ Afterwards, I tried to prove it for $n=k+1$. $$(k+1)!&gt;2^{k+1}$$ $$(k+1)k!&gt;(2)(2^k)$$ Since $k!&gt;2^k$, am I allowed to simply do $$(k+1)(2^k)&gt;(2)(2^k)$$ $$(k+1)&gt;2$$ and this is true for all $k\geq3$. Would the last two steps be illegal because I only assumed that $k!&gt;2^k$ is true?
69440
Show by induction that for all natural numbers n>3  $(1+\frac{1}{n})^n&lt;n$  Let $(1+\frac{1}{n})^n&lt;n$ be true !   We show that $(1+\frac{1}{n+1})^{n+1})&lt;n+1$  $(1+\frac{1}{n})^n(1+\frac{1}{n})&lt;n(1+\frac{1}{n})$   $(1+\frac{1}{n})^{n+1}&lt;n+1$  So now we just need to show that   $(1+\frac{1}{n+1})^{n+1}&lt;(1+\frac{1}{n})^{n+1}$  So here , If I show that :   $(1+\frac{1}{n+1})&lt;(1+\frac{1}{n})$   Does that mean im done with exercise ?
56831
Let $$x_n=\frac{1}{1^2}+\frac{1}{2^2}+\ldots+\frac{1}{n^2},\;\forall n\geq 1.$$ Show that $$\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq \frac{3n-2}{2n-1}.$$   I tried to prove it by induction  but I got a contradiction.
37910
$m,n$ are integers bigger than 0.   I need to prove:  $$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \geq 1$$   I tried to multiply both sides by the common factor and raise both sides to the power of $m*n$ but it did not work for me, and I have no other idea how to proceed.  I don't want the full solution, but just a hint.  Thank you.