Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to find a n(n+1)(2n+1)/6 , but I cannot find it.

$f(x) = \frac{-x^2}4 + 6$ over the points $[0,4]$

The summation inside the brackets is $R_n$ which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

Calculate $R_n$ for the function with the answer being the function of $n$ without any summation signs.

share|cite|improve this question

If I have understood correctly, you want to evaluate: $$A=\lim_{n\to+\infty}\sum_1^nf(x_i)\Delta x, ~~f(x)=\frac{-x^2}{4}+6$$ Set $\Delta x=\frac{4-0}{n}$ in which we divide the closed interval $[0,4]$ into $n$ sub intervals, each of length $\Delta x$. In fact $$x_0=0, x_1=0+\Delta x, x_2=0+2\Delta x,...,x_{n-1}=0+(n-1)\Delta x, x_n=4$$ Since, the function $f(x)$ is decreasing on the interval, so the absolute min value of $f$ on the $i$th subinterval $[x_{i-1},x_i]$ is $f(x_i)$. But $f(x_i)=f(0+i\Delta x)$ as you see above and then $$f(x_i)=-\frac{(i\Delta x)^2}{4}+6=-\frac{i^2\Delta^2x}{4}+6$$

Now, let's try to find the above summation: $$A=\lim_{n\to+\infty}\sum_1^nf(x_i)\Delta x=\lim_{n\to+\infty}\sum_1^n\left(-\frac{i^2\Delta^2x}{4}+6\right)\Delta x$$ wherein $\Delta x=\frac{4}{n}$.

enter image description here

share|cite|improve this answer
So what is Rn = ?? – user59255 Jan 22 '13 at 19:35
I know, I need to know the function. What is it and how do you find it? – user59255 Jan 22 '13 at 19:40
@user59255: Isn't the function you noted, our function? Do you know anything about the Riemannian sum for an definite integral? For finding that summation, we need a continuous function considered on an interval and you have one. $f(x)=6-x^2/4$ – Babak S. Jan 22 '13 at 19:45
So, Rn = 17.84 and lim n--> inf of Rn = 56/3 ?? – user59255 Jan 23 '13 at 3:58
@user59255: yes. see the fig above. – Babak S. Jan 23 '13 at 5:33

The right endpoints of the subintervals are $4k/n$ for $1 \le k \le n$, and the common width of subintervals is $4/n$. So the value of $R_n$, the right hand sum, is $$\sum_{k=1}^n [ (-1/4) \cdot (4k/n)^2 +6]\cdot(4/n)$$ The constant +6 added $n$ times gives $6n$, which is then multiplied by $(4/n)$ so contributes $+24$ to $R_n$. The remaining part of the sum is $$(-1/4)\cdot (4^2/n^2) \cdot (4/n) \cdot \sum_{k=1}^n k^2.$$ From here just use your formula for the sum of the squares of the first $n$ positive integers, plug in and simplify, not forgetting to add the $+24$.

share|cite|improve this answer

Your Answer


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.