Well, just apply induction

Algebra Level 2

$\large \displaystyle\sum_{r=1}^n r = \dfrac{n(n+1)}{2}$

It is well known that the formula for the sum of the first $n$ natural numbers is given as above. Keeping in mind of the formula stated above, is the summation below true?

$\large\displaystyle\sum_{r=1}^n (r+1) = \dfrac{(n+1)(n+2)}{2}$

False True Sometimes

3 solutions

Nihar Mahajan
Jun 14, 2015

I also did this mistake about 2 years ago when I was learning Sigma Notation.

$\Large{\displaystyle\sum_{r=1}^n (r+1) \\ = \displaystyle\sum_{r=1}^n r + \displaystyle\sum_{r=1}^n 1 \\ = \dfrac{n(n+1)}{2} + n \\ = \dfrac{n^2+n+2n}{2} \\ = \dfrac{n(n+3)}{2} \neq \dfrac{(n+1)(n+2)}{2}}$

Moderator note:

Alternative, you can do this:

$\begin{aligned} \sum_{r=1}^n (r+1) &=& 2 + 3 + 4 + \ldots + n + (n+1) \\ &=& (1 + 2+3+\ldots + n) + n \\ & \ne & (1 + 2+3+\ldots + n) + (n+1) = \frac{(n+1)(n+2)}2 \end{aligned}$

(Amended)

It fails for n=1. Enough said.

David Young - 5 years, 8 months ago

True story xD

Mehul Arora - 5 years, 8 months ago

Nice Question Nihar! :D

Mehul Arora - 6 years ago

Thanks!!! :3

Nihar Mahajan - 6 years ago

@Calvin Lin I think there is some flaw in Challenge Master Note because we need to show $\displaystyle\sum_{r=1}^n (r+1) - \displaystyle\sum_{r=1}^n r \neq (n+1)$ rather than showing $\displaystyle\sum_{r=1}^n (r+1) - \displaystyle\sum_{r=1}^n r \neq 0$ , because its obvious that $\displaystyle\sum_{r=1}^n (r+1) \neq \displaystyle\sum_{r=1}^n r$

Nihar Mahajan - 6 years ago

I went about solving it just like the Challenge Master did. Here's how I thought about it:

If $\displaystyle\sum_{r=1}^n r = \frac{n(n+1)}{2}$ , then $\displaystyle\sum_{r=1}^{n+1} r= \frac{(n+1)(n+2)}{2}$

But $\displaystyle\sum_{r=1}^n (r+1)$ can be rewritten as $\displaystyle\sum_{r=2}^{n+1} r$

Clearly $\displaystyle\sum_{r=1}^{n+1} r \neq \displaystyle\sum_{r=2}^{n+1} r$ because it misses the first term.

Finally, because $\displaystyle\sum_{r=1}^{n+1} r= \frac{(n+1)(n+2)}{2}$ and $\displaystyle\sum_{r=1}^{n+1} r \neq \displaystyle\sum_{r=2}^{n+1} r = \displaystyle\sum_{r=1}^n (r+1)$ , we have shown that $\displaystyle\sum_{r=1}^n (r+1) \neq \frac{(n+1)(n+2)}{2}$ .

This is the same work that the Challenge Master did, only I used sigma notation while he expanded it to make it clearer to someone else who was reading it.

Garrett Clarke - 5 years, 12 months ago

Anubhav Sharma
Oct 19, 2015

Rwit Panda
Jun 14, 2015

Simple, r will be summed from 1 to n and 1 will be added n times. So, n(n+1)/2 + n = n(n+3)/2. The sigma symbol indeed confuses.

Well, just apply induction

3 solutions

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