Use mathematical induction to prove that $\forall n\in \mathbb{N}$, $$\sum ^{n}_{i=1}i^{2}=\dfrac {n\left( n+1\right) \left(2n +1\right) }{6}$$
6333
I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:  $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$  I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
115268
Prove by induction: $$\sum_{k=1}^{n} \binom n k = 2^n -1 $$ for all $n\in \mathbb{N}$.   Today I wrote calculus exam, I had this problem given.  I have the feeling that I will get $0$ points for my solution, because I did this:   Base Case: $n=1$ $$\sum_{k=1}^{1} \binom 1 1 = 1 =  2^1 -1 .$$  Induction Hypothesis: for all $n \in \mathbb{N}$: $$\sum_{k=1}^{n} \binom n k = 2^n -1 $$  Induction Step: $n \rightarrow n+1$ $$\sum_{k=1}^{n+1} \binom {n+1} {k} = \sum_{k=1}^{n} \binom {n+1} {k} + \binom{n+1}{n+1} = 2^{n+1} -1$$  Please show me my mistake because next time is my last chance in this class.
67186
I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction.  I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$.  I however have trouble with the second one, I don't know what to start exactly. Any help/hints?
68440
$\displaystyle\sum_{i=0}^n2^i=2^{n+1}-1$.  I don't understand induction so I could use some help.
104668
I need to prove this $$\sum_{k=0}^n \binom{n+k}{k}=\binom{2n+1}{n}$$ Using induction and basically only the Pascal's identity and the symmetry property for binomial coefficients. Now, I noticed that $\sum_{k=0}^n \binom{r+k}{k}=\binom{r+n+1}{n}$ has a relatively simple proof by induction and it basically holds for any $r$. Could I apply that property for $r=n$ and call it a day? I also tried to solve the problem by using a similar argument, the induction step however states that $$\sum_{k=0}^{n+1} \binom{n+1+k}{k}=\sum_{k=0}^{n} \binom{n+1+k}{k}+\binom{2n+2}{n+1}$$ then $$\sum_{k=0}^{n+1} \binom{n+1+k}{k}=\sum_{k=0}^{n} \binom{n+k}{k}+\sum_{k=0}^{n} \binom{n+k}{k-1}+\binom{2n+2}{n+1}$$ Using the induction hypothesis $$\sum_{k=0}^{n+1} \binom{n+1+k}{k}= \binom{2n+1}{n}+\sum_{k=0}^{n} \binom{n+k}{k-1}+\binom{2n+2}{n+1}$$ And that's where I got stuck. Is this a dead end?
96526
I have to show the inequality of $$\left(\sum_{i=1}^n x_i\right)*\left(\sum_{i=1}^n \frac{1}{x_i}\right) \geq n^2.$$For $x_1, ... x_n \in \mathbb{R_{&gt;0}}$ and $ n \geq 1$. I wanted to show this inequality by induction. The basis is clear, but I am not sure how to do the inductive step. Can someone help me with that?
115678
I'm trying to apply an induction proof to show that $((n+1) 2^n - 1 $ is the sum of $(i 2^{i-1})$ from $0$ to $n$.   the base case: L.H.S = R.H.S we assume that $(k+1) 2^k - 1 $ is true. we need to prove that $(k+2) 2^{k+1} - 1$   My try to prove 3 is as follows:  $(k+2) 2^{k+1} - 1$  $(k+2) (2^k * 2) - 1$ , from 2: $2^k = 1/(k+1)$  $(k+2) (2 / (k+1)) - 1$  $(k+1) 2 - 1$  My question is, how could I get to $2^k$ from the last line to prove this formula is right?
117028
I need to use mathematical induction to show that for any n $\in$ $\Bbb N$ ,  if n  $\ge$ 2, then $\prod_\limits{i=2}^n $ $\left( 1-\frac{1}{i^2}\right)$ = $\frac{n+1}{2n}$
85800
I'm having trouble proving this by induction. We need to show that $P(k+1)$ is true: $$\sum_{i=2}^{k+1} \frac{1}{i^2}\leq 1.$$  Don't know where to go from here. Any help?
122652
I need to prove $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$ using induction. I am getting stuck midway through the inductive step.   Here is what I have:  $\forall n\geq 2$, where $n\in\mathbb{N}, P(n)$ is the statement "$\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$".  Base case:  $\sum\limits_{i=2}^{2}\frac{1}{(i-1)i}$ = $\frac{1}{2}$ = $\frac{(2-1)}{2}$ so $P(2)$ is true.  Inductive assumption: Assume $P(n)$ is true. Need to show $\sum\limits_{i=2}^{n+1}\frac{1}{(i-1)i}$ = $\frac{((n+1)-1)}{(n+1)}$.  $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ + $\frac{1}{n^{2}}$ = $\frac{(n-1)}{n}$ + $\frac{1}{n^{2}}$  = $\frac{n(n-1) + 1}{n^{2}}$ = $\frac{n^{2}-n+1}{n^{2}}$  I got stuck here and tried working backward from $\frac{((n+1)-1)}{(n+1)}$, which gives me:  $\frac{((n+1)-1)}{(n+1)}$ = $\frac{n}{(n+1)}$   Any suggestions on what to do next or where I may be going wrong?
108482
How can the following be proved by induction?  $$\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$$ I am out of ideas after practicing for a while:  $$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{2n-1}-\frac{1}{2n}\right) $$  Does this involve telescoping series?
104310
I need to prove that for every natural n that happens and the prove needs to be by induction, I wont prove for n=1 because its obvious,so lets move to n and n+1 $$ \frac{(2n!)}{(n!)^2} ≥ \frac{(4^n)}{(2n+1)} $$ the assumption for n+1 $$ \frac{(2n+2)!}{(n+1)!^2} ≥ \frac{(4^{n+1})}{2n+3} $$ now I dont now how to progress,I tried this for n+1 $$ \frac{(2n+2)(2n+1)(2n!)}{(n+1)!^2} ≥ \frac{(4^{n}4^1)}{(2n+1)+2)} $$ and had a suggestion to use $ 2n! = (n!) \prod _{n=n}^{2n}\:n\cdot )! $ but cant see how it helps,ty.
37046
I met an inequality, I ask, do not mathematical induction to prove that:     Prove \[ \left(\frac n2\right)^n > n! > \left(\frac n3\right)^n \] without using induction
63335
Prove that $1&lt;\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.  By using the Mathematical induction. Suppose the statement holds for $n=k$.  Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}=(\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1})+(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$  we know $\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}&gt;1$  What can we do for $(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$?
102725
Prove $\sum_{i=1}^n2^{i-1}=\sum_{i=0}^{n-1}2^i=2^n-1$ combinatorially.  This is easy to prove inductively.  I know that $\sum_{i=0}^n{n\choose i}=2^n$ so maybe change $\sum_{i=0}^{n-1}2^i$ to $\sum_{i=0}^{n-1}\sum_{j=0}^i{i\choose j}$.  I also know that $2^n-1$ is the number of proper subsets in a set of size $n$.
117647
Let n be a positive integer.  Prove that $$\sum_{i=1}^n 2i\binom{2n}{n-i}= n\binom{2n}{n}.$$
37322
Prove using induction that $\forall n\in\mathbb N, \forall x\in \mathbb R: 0&lt;x&lt;1: (1-x)^n&lt;\frac 1 {1+nx}$   My attempt:   Base: for $n=1: 1-x&lt;\frac 1 {1+x}\iff 1-x^2&lt;1$, true since $0&lt;x&lt;1$.  Suppose the statement is true for $n$, prove for $n+1$:  $(1-x)^{n+1}=(1-x)(1-x)^{n}\overset{i.h}&lt;\frac{(1-x)}{1+nx}$  Now I got stuck, maybe another induction to show that $1+nx+x&lt;1+nx$? Is there another way?  Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $\frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?