I've considered the equation by modulo 9 and tried the binomial theorem  $$4^n=(6-2)^n$$ but I still cannot prove it. Maybe there is a clever solution but so far I have been unable to spot it. Can anyone help me?
82351
Prove that for all $n\in\mathbb N$, $9 \mathrel| (4^n+6n-1)$.  I have shown the base case if $n=0$ (which is a natural number for my course).  I'm now trying to prove that $9k=4^n+6n-1$.  I substituted $n+1$ for $n$, and have $4^{(n+1)}+6(n+1)-1$, which equals $4*4^n+6n+5$.  I'm stuck. Is that not the right way to go? I don't know how to get from the “$+5$” to “$-1$” so I can substitute $9k$ to the right side of the equation. Any help would be greatly appreciated.
49298
I'm trying to prove that  \begin{equation} \sum_{k= 0}^{n} k \binom{n}{k} = n \cdot 2^{n - 1} \end{equation}  with the Binomial Theorem.  I know that the B.T. states that  \begin{equation} (x + y)^n =  \sum_{k= 0}^{n} \binom{n}{k} x^{n-k}y^{k}  \end{equation}  The proof that I have tried starts with:  \begin{equation} n \cdot 2^{n-1} =  \end{equation}  Actually I am stucked with this step.
123923
I'm trying to prove binomial theorem by induction, but I'm a little stuck. I would look at online resources as this problem has been done many times, but the version I am trying to prove the binomial theorem in a different form.  $$(1 + x)^n = \sum_{k = 0}^{n} \binom{n}{k} x^k$$  I'm mostly confused as to how I can make the left side be equivalent to a summation, any help is appreciated. Try to hint me along!
63201
Find all real solutions to the equation:$$4^x+6^{x^2}=5^x+5^{x^2}$$  Obviously $0$ and $1$ satisfy this equation, but here $x$ is a real number, and I do not know how to do the next step, in fact, I guess this equation is not in addition to $0$ and $1$ other than the root. But you know, I need a proof! Could anyone help me? Thanks a lot!
17558
I've done the base step, but how do I prove it is true for $n+1$ without using a fallacy? $$4(n!)&gt;2^{n+2}\quad \text{for } n\geq 4$$ Please help.
23341
Prove that the number $14641$ is the fourth power of an integer in any base greater than $6$?  I understand how to work it out, because I think you do   $$14641\ (\text{base }a &gt; 6) = a^4+4a^3+6a^2+4a+1= (a+1)^4$$  But I can't understand why they had to specify that the base is greater than 6? Is that because if it's 4, then 4 will be cancelled out in the equation and if it's 6, it will be too?   Please advise.  Sorry for asking such a trivial question.
91929
I had recently faced a problem:     Solve the Diophantine Equation $x^2 - y! = 2001$.   Solving it was quite easy. You show how $\forall y \ge 6$, $9|y!$ and since $3$ divides the RHS, it must divide the LHS and if $3|x^2 \implies 9|x^2$ and so the LHS is divisible by $9$ and the RHS is not. Contradiction. Hence, the only solution is $(45, 4)$.  That made me wonder, how we can solve the Diophantine Equation $x^2 - y! = 2016$. We cannot apply the same logic here. $2016$ is a multiple of $9$ and it is clear that $3|x$ and $9 \nmid x$. How do I proceed from here?
83623
I'm trying to prove the following equation above. So far I have: \begin{align} 2^{2n} &amp;= (1+1)^{2n}\\ &amp;= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{n-k} = \sum_{k=0}^{2n}\binom{2n}{k} &amp; \text{(By the Binomial Theorem)} \end{align}  I know I have to use the following identity somehow: $$\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$$  How do I split my summation to get what I'm looking for? Thanks!  EDIT: HERE IS MY SOLUTION \begin{align*} 2^{2n} &amp;= (1+1)^{2n}\\ &amp;= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{2n-k} &amp; \text{(By the Binomial Theorem)}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{k}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{2n-k} &amp; \text{(Binomial Symmetry)}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=0}^{n-1}\binom{2n}{k}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\ &amp;= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\ &amp;= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n+1}{k} &amp; \text{(By Identity listed above)}\\ &amp;= \sum_{k=0}^{n}\binom{2n+1}{k} \end{align*}
26130
Why is $\sum_{k=0}^n (_k^n) 2^k$ simplified as $3^n$ and not just $2^n$ ? The answer uses the binomial theorem as a solution: let $a=1$ and let $b=2$. So $1+2 = 3$. But since $1+2 = 3$ why do we not use $1+1 = 2$?
17521
Examining the Tartaglia's triangle, I have observed that all the squares were the trivial cases, that is, $\binom{n^2}1$ or $\binom{n^2}{n^2-1}$.  More formally:  Conjecture: If $\binom nm=k^2$ then $n=k^2$.  Is it known to be true?  I have tried to use the formula  $$\nu_p\left(m!\right)=\sum_{k=1}^\infty\left\lfloor \frac m{p^\alpha}\right\rfloor$$  to prove that the exponents of the factorization of the binomial coefficients are odd, but I realized that this cannot be proved, because the binomial coefficients needn't be square-free: $\binom 63=20$, for example.  Any ideas?
62005
4 . (a) Prove that   $$\sum_{k=0}^l \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}.$$   Hint: Apply the binomial theorem to $(1+x)^n(1+x)^m$.   I'm having a hard time trying to solve the problem above. I've done all of the previous exercises of the 2nd chapter with little difficulty, so far. I think I might be missing a trivial point somewhere.    The answer I got from the Answer Book, and is not very helpful either... :(      4. (a) Since $(1+x)^n(1+x)^m = (1+x)^{n+m}$  we have  $$\sum_{k=0}^n \binom{n}{k}x^k\cdot\sum_{j=0}^m \binom{m}{j}x^j\cdot=\sum_{l=0}^{n+m} \binom{n+m}{l}x^l$$  But the coefficient of $x^l$ on the left is clearly $$\sum_{k=0}^l\binom{n}{k}\binom{m}{l-k}.$$  One term of the sum occurring for each pair $k$, $j = l-k$.     I couldn't get the last part of the answer: why is it that the "coefficient of $x^l$ on the left is clearly $\sum_{k=0}^l\binom{n}{k}\binom{m}{l-k}$."?
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?
81459
I'm trying to prove that $6 \mid (n^3 - n)$ where $n$ is a nonnegative integer. I started off by proving the basic step with $P(6)=4$. The next step would be the induction. However I'm having a bit f trouble understanding and using the method. Could someone show me how I would prove the statement and explain to me how I would get the answer. Thanks in advance.
9540
Fix integer $n\geq 1$. Consider the number $$1-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots$$where the sum continues as long as the lower number in the binomial is $\leq n$. Is there a way to simplify this sum?   The first few values are $1, 0, -2, -4, -4$.
31421
Why should we expect that $$2^n=\sum_{k=0}^n {n\choose k}$$   It is easily seen to be true, by the binomial theorem:  just set $x=y=1$ in $(x+y)^n$.  But what is an intuitive reason why it is true (in terms of subsets)?
27556
Could we use the binomial theorem for non integers? This comes from:   $$\sqrt{(a+b)}$$ which I can write as $$(a+b)^{1/2}$$  Could I then use the binomial theorem to figure out the value of this expression?
60296
I am a CS undergrad and I'm studying for the finals in college and I saw this question in an exercise list:     Prove, using mathematical induction, that $2^n &gt; n^2$ for all integer n greater than $4$   So I started:  Base case: $n = 5$ (the problem states "$n$ greater than $4$", so let's pick the first integer that matches)  $2^5 &gt; 5^2 \implies 32 &gt; 25$ - ok!  Now, Inductive Step:  $2^{n+1} &gt; (n+1)^2$  now expanding  $2 * 2^n &gt; n^2 + 2n + 1$  Well, I can't go on, I don't know how to finish this proof. Could anyone help me with this one?  Also, I know that the statement (inductive hypothesis) is true. Simply, I'll have a number to an exponent $n$ twice, next to $n$ squared and $2n$ (which could be seen as quadratic and linear functions) which "grows" much lower than the exponential one. It could be easily "proved by example" with any $n$ greater than $4$.  I want to know how to prove it algebraically , so my professor does not freak out.  thanks in advance.