Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am quite lost on this question:

(a) For $n\in \mathbb{N}$, use induction to show that

$$\sum_{k=1}^{n}k^{5}=\frac{2n^{6}+6n^{5}+5n^4-n^2}{12}$$

(b) Fix $b>0$. Use the definition of the definite integral together with the Riemann partition $$P_{n}:=\left \{ (x_{k},\left [ x_{k-1},x_{k} \right ]) \right \}$$ where $x_{k}:=\frac{kb}{n}$ for $0\leq k\leq n$, to compute $\int_{0}^{b}x^{5}dx$

I know with induction that I should be starting with a base case, like $n=1$. My textbook has 1-2 pages about induction but I am not able to properly understand and apply it here. Part (a) must help with the solving of part (b)

share|improve this question

2 Answers 2

up vote 1 down vote accepted

(a): To use induction, you need to prove the equation for the base case (in this case $n=1$), then prove that given the equation is true for $n=m$, it is true for $n=m+1$ as well. In other words, you show that (eqn. is true for $n=m$) implies (eqn. is true for $n=m+1$). Then, since the equation is true for $n=1$, it must be true for $n=2$ ($m=1$ and $m+1=2$). By the same logic, it is true for 2, 3, 4, etc., and therefore it is true for all $n\in\mathbb{N}$.

Here, the base case is clearly true: the left hand side is 1 and the right hand side is $\frac{2+6+5-1}{12}=1$. For the inductive step, you know that $\sum^{m+1}_{k=1} k^5=\sum^{m}_{k=1} k^5+(m+1)^5$. So you need to show that $[2(m+1)^6+6(m+1)^5+5(m+1)^4-(m+1)^2]/12=(2m^6+6m^5+5m^4-m^2)/12+(m+1)^5$. Unfortunately, this requires a lot of messy algebra.

(b): To integrate a function $f$, you need to add up all the $f(x_k)(\Delta x)_k$'s. (For each sample rectangle $k$, $f(x_k)$ is the value of $f$ somewhere in the rectangle, and $(\Delta x)_k$ is the width of the rectangle). In this case, $f(x)=x^5$ and $x_k=kb/n$, so $f(x_k)=(b/n)^5 * k^5$. You are cutting up the interval $[0,b]$ into $n$ rectangles, so the width of each one is $b/n$. The Riemann sum is $\sum^n_{k=1} f(x_k)(\Delta x)_k=\sum^n_{k=1} (b/n)^5 * k^5 * (b/n)=(b/n)^6 \sum^n_{k=1} k^5=(\frac{b}{n})^6 \frac{2n^6+6n^5+5n^4-n^2}{12}=b^6 \frac{2n^6+6n^5+5n^4-n^2}{12n^6}$. The integral is the limit of the Riemann sum as the number of sample rectangles ($n$) goes to infinity. The $n^5,n^4,n^2$ terms become negligible compared to the $n^6$ term as $n$ becomes very large, so the final answer is $\frac{1}{6}b^6$.

share|improve this answer

You are being asked (in (a)) to prove a statement about an integer $n$. As you say, you first establish it for $n=1$. Then you write down the statement for $n$, and you write down the statement for $n+1$, and you show that you can deduce the second statement from the first. By induction, that will prove the statement for all (positive integers) $n$. Have a go at that and then we'll get back to the integral.

share|improve this answer
    
$$\sum_{k=1}^{n}k^{5}=1^{5}+2^{5}+3^{5}+\cdots +(n-1)^{5}+n^{5}$$ $$\sum_{k=1}^{n+1 }k^{5}=1^{5}+2^{5}+3^{5}+\cdots +n^{5}+(n+1)^{5}$$ $$=\frac{2n^{6}+6n^{5}+5n^4-n^2}{12}+(n+1)^{5}$$ $$=\frac{2n^{6}+6n^{5}+5n^4-n^2+12(n+1)^{5}}{12}$$ –  Malthus Nov 24 '11 at 6:49
    
$$=\frac{2n^{6}+18n^{5}+65n^{4}+120n^{3}+119n^{2}+60n+12}{12}$$ The last is as far as I am able to go. I cannot figure out how to manipulate it into: $$=\frac{2(n+1)^{6}+18(n+1)^{5}+65(n+1)^{4}+120(n+1)^{3}+119(n+1)^{2}+60(n+1)+12‌​}{12} $$ –  Malthus Nov 24 '11 at 7:06
    
That last expression is not where you want to go. You want to go to the statement for $n+1$, which is the equation in your question but with $n$ replaced everywhere by $n+1$. –  Gerry Myerson Nov 24 '11 at 12:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.