My approach is to solve this by induction.   Base case: $n=1$  $1^3 = 1^2 = 1$  Inductive Step: Suppose that  $1^3+2^3+...+n^3 = (1+2+...+n)^2$  holds for all positive integers $n$. We use that to show that  $1^3+2^3+...+n^3 + (n+1)^3 = (1+2+...+n+(n+1))^2$   $1^3+2^3+...+n^3 + (n+1)^3 = (1+2+...+n)^2 + (n+1)^3$  And here is where I can no longer proceed. I know there must be some trick/computation that I am not seeing that will make this proof easier. Help? Thanks!!
1911
How can I prove that     $$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$    for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.  Thanks
93701
Question: Prove $\forall n\geq 2,n\in\mathbb{Z}$, $(n+1)\mid(n^3+1)$    I know that it is possible to solve by factoring $n^3+1$ and showing that $n+1$ is a multiple, but I would like to show this via induction.  Attempt: Using induction:  Let $n=2:3\mid 9$, so true for $n=2$  (1) Assume true for $n=k:(k+1)\mid(k^3+1)$, so $(k+1)r=k^3+1$ for some $r\in\mathbb{Z}$  (2) Let $n=k+1:(k+2)\mid(k+1)^3+1$, so $(k+2)s=(k+1)^3+1$ for some $s\in\mathbb{Z}$  From (2) we have $(k+1)s+s=(k+1)^3+1$, subbing (1) into (2), I get $k^3+1+s=(k+1)^3+1$  Therefore $k^3+s=(k+1)^3$.  Setting $s=3k^2+3k+1$, this holds.  Where am I going wrong, is the proof valid, any help would be greatly appreciated.
46321
The question asks to verify that each equation is true for every positive integer n.   The question is as follows:  $$1+ 3 + 5 + \cdots + (2n - 1) = n^2$$  I have solved the base step which is where $n = 1$.  However now once I proceed to the inductive step, I get a little lost on where to go next:  Assuming that k is true (k = n), solve for k+1:  (2k - 1) + (2(k+1) - 1) (2k - 1) + (2k+2 - 1) (2k - 1) + (2k + 1)   This is where I am stuck. Do I factor these further to obtain a polynomial of some sort? Or am I missing something?
25221
Prove by induction that for all $n \ge 0$:  $${n \choose 0} + {n \choose 1} + ... + {n \choose n} = 2^n.$$  In the inductive step, use Pascal’s identity, which is:  $${n+1 \choose k} = {n \choose k-1} + {n \choose k}.$$  I can only prove it using the binomial theorem, not induction.
63872
I am currently working on this problem from Hardy's Course of Pure Mathematics and have gotten stuck near the end. I was wondering if someone could help me determine what to go next.  Question  If $a_1, a_2, ...,a_n$ are all positive and $S_n=a_1+a_2+...+a_n$ then:  $(1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}$  My Attempt  Proof by induction:  I had first shown that it was true for $n=1$ and $n=2$.  Now suppose that it is true for n. Then it must also be true for $n+1$  $(1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1}) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!} + \dfrac{(S_{n})^{n+1}}{(n+1)!}$  Define: $(1+a_1)(1+a_2)...(1+a_n)=x$ and $1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}=y$  Then: $x+xa_{n+1} \le y+\dfrac{(S_{n})^{n+1}}{(n+1)!}$  By the induction assumption, we know that $x \le y$.  What I can't figure out  From this point, what I think the next natural step would be is to show that  $xa_{n+1} \le \dfrac{(S_{n})^{n+1}}{(n+1)!}$  I know this rearranges to:  $xa_{n+1} \le \dfrac{(S_{n})^{n}}{n!} \times \dfrac{S_n}{(n+1)} $  And feel there may be something I can do here, but haven't been able to figure anything out.
114927
I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!     Problem:      For any natural number $n , n^3 + 2n$ is divisible by $3.$   This makes sense      Proof:      Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$   $2 \times 0 = 0.$ So it is divisible by $3.$      Induction: Assume that for an arbitrary natural number $n$,   $n^3+ 2n$ is divisible by $3.$       Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use   the induction hypothesis. Got it      $$( n + 1 )^3+ 2( n + 1 ) = (  n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$      $$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying  and regrouping}\}$$   $$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out the 3}\}$$      which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$   by the induction hypothesis. What?   Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$
36259
Prove that $2^{2\cdot 3^{n-1}}\equiv 1+3^n\pmod{3^{n+1}}$ for every natural $n$   Of course, should be done by induction. Base case ($n=1$) is easy.  I got stuck with the step:  Let's assume that it's right for $n$. Then,  $$2^{2\cdot 3^{n-1}} = 1 + 3^n \pmod{3^{n+1}}$$  We want to evaluate $$2^{2 \cdot 3^n} \pmod {3^{n+2}}$$  I think we can somehow utilize the fact that $2,3$ are coprime and reduce $3^{n+2}$ to $3^{n+1}$.   I'd be glad for help on that.  Thanks!
58071
So I'm trying to use mathematical induction to show that for all integers $n \ge 1$ ,  $$ 8|(3^{2n} - 1)$$  (is divisible by 8)  I have my base case: [P(1)],  $3^2 - 1 = 9 - 1 = 8$, since $8|8$, the base case proves true  Assume [P(k)], $ 8 | (3^{2k} - 1)$.  I know that I need to show [P(k+1)], $ 8| (3^{2(k+1)}-1) $, but I'm not sure how to prove this. I've only been using induction for summation, so how could I prove divisibility?
851
$ 169$ | $3^{3n+3}-26n-27$ ?  Fulfilled for $n=0$. Induction to $n+1$: An integer $x$ exists so that  $ 169x= 3^{3n+6}-26n-27-26$ $ 169x= 27*3^{3n+3}-26n-27-26$ $ 169x= 26*3^{3n+3}+3^{3n+3}-26n-27-26$  An integer $m$ exists so that   $ 169x= 26*3^{3n+3}+169m-26$ ($ 13x= 2*(3^{3n+3}-1)+13m$)  Now I'm stuck although it looks simple. Thanks for any input in advance.
28654
I want to prove that $2^{n+2}+3^{2n+1}$ is divisible by 7 using induction. My first step is replace $n$ with $1$.   $2^{1+2}+3^{2(1)+1}$ $2^3+3^3$ $8+27$ $35 = 7\times 5$   The next step is assume that $2^{n+2}+3^{2n+1}$ is divisible by 7. And the last step is to prove that $2^{(n+1)+2}+3^{2(n+1)+1}$.   $2^{(n+1)+2}+3^{2(n+1)+1}$ $2^{n+3}+3^{2n+3}$ I got stuck here.   How can I prove it using induction?
12895
So what I am trying to prove is for any real number x and natural number n, prove $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$   I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis. My base case is when $n=2$ we have on the left side of the equation $x^2-1$ and on the right side: $(x-1)(x+1)$ which when distributed is $x^2-1$. So my base case holds.   Now I assume that $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ for some $n$. However, this is where I am stuck. Am I trying to show $x^{n+1}-1=(x-1)(x^n + x^{n-1}+x^{n-2}+...+x+1)$? I am still a novice when it comes to these induction proofs. Thanks
101831
I am trying to prove that for every element of $\mathbb{N}$, that                    $2^n \leq 2^{n+1} - 2^{n-1} - 1.$ I started by showing that initial case, of $n=1$, is true. Then I proceed to the inductive step by assuming that for some $k$ in $\mathbb{N}$,                     $2^k \leq 2^{k+1} - 2^{k-1} - 1$.  Then, for $n=k+1$,                    $2^{k+1} \leq 2^{k+2} - 2^{k} - 1$. Is this sufficient to prove the statement by induction or do I have to expand the exponentials to rearrange the inequality in a different way?
6512
Prove that if $n&gt;1$ is odd, then $1^n+2^n+3^n+\ldots+(n-1)^n$ is divisible by $n$.   I'm pretty sure I need to prove this using induction but I am unsure as to how to go about it.
120942
This is also a question on my exam paper that i proved by using mathematical induction. However, my tutor tells me that it can be proved without using mathematical induction. I really want to know how to deal with in another way.  Show that for any positive integer n, $(3n)!/(3!)^n$ is an integer.
8761
I'm working on a task where I'm a bit unsure if the answer I've got is correct.  Here is the task:     Show by induction that the following assertion is true for all natural   numbers $n$      $n^3 - n$ is divisible by $3$   Here is my answer:     For $n = 1$,   $n^3 - n = 1 - 1$   which is divisible by $3$      Assume the statement is true for some number $n$, that is, $n^3 - n$   is divisible by $3$. Now,      $(n + 1)^3 - (n + 1) = n^3 + 3n^2 + 3n + 1 - n - 1 = (n^3 - n) + 3(n^2 + n) = (n^3 - n) + 3n(n+1)$      which is $n^3 - n$ plus a multiple of $3$.      Since we assumed that $n^3 - n$ was a multiple of $3$, it follows that   $(n + 1)^3 - (n + 1)$ is also a multiple of $3$.      So, since the statement "$n^3 - n$ is divisible by $3$" is true for $n  = 1$, and its truth for $n$ implies its truth for $n + 1$, the statement is true for all whole number $n$.   I would appreciate if someone could go through the task and the answer and see if I've done this correctly.  Thanks a lot!
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.
49946
How do I prove the following statement by induction?  $$n^2 \lt 2^n$$  $P(n)$ is the statement $n^2 \lt 2^n$  Claim: For all $n \gt k$, where $k$ is any integer, $P(n)$  (since $k$ is any integer, I assume I have to prove this for positive and negative integers)  So let base case be $P(1)$, and I have to prove $P(n)$ for $n \ge 1$ and $n \lt 1$  $P(1)$ is $1^2 \lt 2^1$ which is clearly true.  Induction hypothesis: $k^2 \lt 2^k$  Inductive step $(k+1)^2 \lt 2^{(k+1)}$  $(k+1)^2 = k^2 + 2k + 1$  $k^2 + 2k + 1 \lt 2^k + 2k + 1$ by inductive hypothesis  Not sure how to proceed. Is my previous intuition that I have to prove this for $n \ge 1$ and $n \lt 1$ correct?