I encountered the following induction proof on a practice exam for calculus:  $$\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$$  I have to prove this statement with induction.  Can anyone please help me with this proof?
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?
75589
Given a positive integer $n$. How to prove that $\sum\limits_{k=0}^{n}(-1)^k \dbinom{n}{k} = 0 $ ?   I've tried using mathematical induction, then:  $$p(1)=\sum_{k=0}^{1}(-1)^k \binom {1}{k}=0 $$  And my induction hypothesis is: $$p(n)=\sum_{k=0}^{n}(-1)^k \binom {n}{k}=0 $$ So, i need prove: $$p(n+1)=\sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k}=0 $$ Let $\sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k}$, $$\begin{align} \sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k} &amp;= \sum_{k=0}^{n}(-1)^{k} \binom {n}{k}+(-1)^{k} \binom {n+1}{n+1} \\ &amp;=0 + (-1)^{k} \end{align}$$ What am i doing wrong?
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I am stuck with the identity  $$ F_{2n} = \sum_{k=1}^n \binom{n}{k} F_k, $$  which happens to be formula . I am using induction, but so far without too much result.  $$ \sum_{k=1}^{n+1} \binom{n+1}{k} F_k = \\ F_{n+1} + \sum_{k=1}^n \binom{n+1}{k} F_k = \\ F_{n+1} + \sum_{k=1}^n \binom{n}{k} F_k + \sum_{k=1}^n \binom{n}{k-1}F_k =\\ F_{n+1} + F_{2n} + \sum_{k=1}^n \binom{n}{k-1}F_k, $$   I am stuck with the sum $ \sum_{k=1}^n \binom{n}{k-1}F_k$, because, on the other hand   $$ F_{n+1}+ \sum_{k=1}^n \binom{n}{k-1}F_k = F_{2n+1} $$  Which would conclude the proof. What am I missing?
101647
I'm working on proving the following statement using induction:  $$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$  Fair enough. I'll start with the basis step:  Basis Step: (n=1) $$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$ $$ \frac{1}{1^2} \le \frac{2}{1+1} $$ $$ 1 \le 1  \checkmark  $$  Inductive Step: $$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$ $$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{n+2} $$  My goal is to prove $\forall_{n\ge1} s(n) \implies s(n+1) $ or that this inequality holds true for all $n\ge1$.  I'm not quite sure to go from here on the inductive step. I understand that I need to basically work some clever substitution and manipulation into the problem to end up with: $$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n}{n+1} $$  However, I'm not quite sure what needs to done to obtain this after attempting a few times.
36530
Prove by induction the summation of $\frac1{2^n}$ is greater than or equal to $1+\frac{n}2$.  We start with $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$$ for all positive integers.   I have resolved that the following attempt to prove this inequality is false, but I will leave it here to show you my progress. In my proof, I need to define P(n), work out the base case for n=1, and then follow through with the induction step. Strong mathematical induction may be used.  This is equivalent to $$\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n}2\;.$$  Let $P(n)$ be summation shown above.   Base case for $n=1$, the first positive integer,   $$\sum_{k=0}^1\frac1{2^k}=\frac1{2^0}+\frac1{2^1}=1+\frac12=\frac32\ge 1+\frac12=\frac32\;,$$  so base case is true.  Induction step: Assume $P(n)$ is true and implies $P(n+1)$. Thus   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge\frac1{2^{n+1}}+\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n+1}2\;.$$  This can be written as   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge \frac1{2^{n+1}}+1+\frac{n}2\ge 1+\frac{n+1}2\;.$$  I work the math out but I get stuck contradicting my statement. Please show your steps hereafter so I can correct my mistakes.
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.
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 $(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$?
87769
Given the following identity: $$\sum_{i=0}^k (-1)^i \binom ni = (-1)^k \binom {n-1}{k}.$$  This is provable by induction. However, I wonder if there is a way to prove this in a combinatorial fashion (something like subsets, groups, choosing $n$ balls from a box with $k$ balls, et cetera.)  Thank you.
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?
26124
Can someone look at my proof. I am supposed to prove by induction. The question is to prove the following:  $$\sum _{i=0}^{n}{i} =\frac { n\left( n+1 \right)  }{ 2 } .$$  If $n=1$  Then  $$1=\sum_{i=1}^1{i}=\frac{1(1+1)}{2}=1$$  Now assume $n=k$.  Thus  $$\sum_{i=1}^k{i}=\frac{k(k+1)}{2}=1$$  If now $n=k+1$  $$\sum_{i=1}^{k+1}{i}=\sum_{i=1}^k{i}+(k+1)= \frac{(k+1)(k+2)}{2}$$  Is there anything more that I have to do, or is this it?
14700
I oversaw a high-school mathematics test the other day, and one of the problems was the following     Show, using induction or other means, that $$\sum_{i = 1}^n\frac1{i(i+1)} = 1-\frac1{n+1}$$   The induction proof is very standard, where the induction step relies on the fact that $\frac{1}{n+1} + \frac{1}{n(n+1)} = \frac{1}{n}$, and I'm sure it's been answered on this site before. However, I got intrigued by the "or other means" part of the question.  I don't know whether the teacher who wrote the test even considered any alternative solutions (he may just have written it so that if anyone has a crazy idea that works out, then they should get a full score for it), but I tried to find one. For instance, we may do the following telescope-ish argument: $$ \frac{1}{1(1+1)} + \frac{1}{2(2+1)} + \cdots + \frac{1}{(n-1)n} + \frac{1}{n(n+1)} + \frac{1}{n+1}\\ = \frac{1}{1(1+1)} + \frac{1}{2(2+1)} + \cdots + \frac{1}{(n-1)n} + \frac1n\\ \vdots\\ = \frac{1}{1(1+1)} + \frac12\\ = 1 $$ However, I feel that this is just an induction proof in disguise (hidden in the vertical dots). If one uses the mechanics of the induction proof to check whether the formula is true for a specific $n$, then one certainly does the exact same calculations as I have done here.  Is there a proof of this fact that clearly does not use induction (or at least hides it better)? The more elementary the better, and the ultimate goal would be to do it within the syllabus of the students taking the test (or at least not far from it). For reference, they should be familiar with the summation formula of arithmetic and geometric series and their derivations (so techniques resembling those would be well within specifications).  If there is a solution using calculus, then the students should be able to integrate elementary trigonometric functions, as well as exponential functions, logarithms and rational functions. They are familiar with integration by parts, substitution and partial fractions.  I welcome more advanced solutions as well, of course.
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I'm finding a closed form to  $\sum^{n}_{k=1} \frac{k}{(k+1)!}  (n \leq 1) $ (in a environment of induction and recurrence)   I've been trying to solve it without success, can anybody help me (?)  The last thing I've tried to do is write the expression as a Non-Homogeneous Linear Recurrence:   $ f(1) = \frac{1}{2} $   but, actually I have no idea of how to attach it.   $ f(n+1) = f(n) + \frac{1}{(n+2)n!} $
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Does anyone have any idea how to prove that  $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}$$
64537
Prove that for every positive integer $n \ge 2$ $$\sum^n_{k=1}k \sqrt{\binom nk}\leq\sqrt{2^{n-1}n^3}$$ I tried it by induction but I didn't know how to end it.
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I am wondering how to prove both of this by induction: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2^n}\leq 1+\frac{n}{2}$ In general, how should I think of how to prove such an inductive step for these types of inequalities? If there is any other way to prove please feel free. Any help would be appreciated.
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I know that $$\sum_{k=0}^n \binom nk k^2=2^{n-2}(n^2+n),$$ but I cannot find a way how to prove it. I tried induction but it did not work. On wiki they say that I should use differentiation but I do not know how to apply it to binomial coefficient.  Thanks for any response.  (Presumptive) Source: Theoretical Exercise 1.12(b), P18, A First Course in Pr, 8th Ed, by S Ross
35245
The exact question:  Prove:  $\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}}\gt2(\sqrt{n+1}-1)$  I have looked at similar problems but still don't understand how to prove this inequality by induction. So far I have this:  Induction basis:  Let n=1  $\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}} = \frac{1}{\sqrt{1}} = 1 &gt; 2(\sqrt{1+1}-1) = ~.828$  $1&gt;.828$  So it proves the inequality true when n=1.   Now i really don't know how to continue even with all the examples i have browsed through. One of them i came across showed that the induction hypothesis should let P(n) equal the equation above and do something with P(n+1). I am not looking for the answer I just need help on how to continue with the problem. What other steps are necessary for me to complete this proof by induction.