# Prove $\sum_{i=n}^{2n}i=\frac{3}{2}n(n+1)$ using induction

Prove that $$\sum_{i=n}^{2n}i=\frac{3}{2}n(n+1)$$ usind induction.

Base step: for $$n=1$$ we have $$\sum_{i=1}^{2\cdot1}i=1+2=3$$, and $$\frac{3}{2}1(1+1)=\frac322=3$$, so it is true for $$n=1$$.

My problem starts when trying to prove the inductive step: $$\sum_{i=h}^{2h}i=\frac{3}{2}h(h+1)\implies\sum_{i=h+1}^{2(h+1)}i=\frac{3}{2}(h+1)(h+2).$$ I have read that it has to be: \begin{align}\sum_{i=h+1}^{2(h+1)}i&=\sum_{i=h+1}^{2h+2}i\\ &=\underbrace{\sum_{i=h}^{2h}(i)}_{\text{Hypothesis}}+\underbrace{\sum_{i=1}^{2}}_{(*)}\\ &=\frac{3}{2}h(h+1)+\underbrace{\sum_{i=1}^{2}}_{(*)}\\ &=\frac{3}{2}h(h+1)+\bigl((h+1)+2(h+1)\bigr)\\ &=\cdots\\ &=\frac32h^2+\frac92h+3\\ &=\frac32(h+1)(h+2)\end{align}

but I think $$(*)$$ has no sense.

But I am not able to write it in a nice way because I think it has to be: $$(*)=\sum_{i=h+1}^{2}(2i+2)=\bigl(2(h+1)+2\bigr)+\bigl(2(2(h+1)+2)\bigr),$$ which is not right.

So my questions are:

1. Is $$(*)$$ correct? What has to be inside the summation i.e. $$\sum_{i=1}^2\square$$? Why the initial value is $$i=1$$ but not $$i=h+1$$?
2. Where is my mistake?

Thanks!!

\begin{align*} \sum_{i=h+1}^{2h+2}i & = \sum_{i=h+1}^{2h}i+\underbrace{(2h+1)+(2h+2)}_{\text{ the } 2h+1 \, \& \, 2h+2 \, \text{terms}} \\ & = \color{red}{h}+\sum_{i=h+1}^{2h}i+(2h+1)+(2h+2)\color{blue}{-h}\\ & = \underbrace{\sum_{i=\color{red}{h}}^{2h}i}_{\text{hypothesis expression}}+(2h+1)+(2h+2)\color{blue}{-h}\\ &=\frac{3}{2}h(h+1)+3h+3. \end{align*} Now you can finish off.

Hint: $$\sum_{k = n + 1}^{2n + 2} k = \left(\sum_{k = n}^{2n + 2} k\right) - n = \left(\sum_{k = n}^{2n} k\right) + \left(2n + 1 + 2n + 2\right) - n = \left(\sum_{k = n}^{2n} k\right) + 3n + 3.$$ Explanation: In the very first expression my first summand is the $$n + 1$$-th term. After the equality sign I begin at the $$n$$-th term, which is not in the first expression, so I have to subtract it again.

For the next equality, it's the same procedure, only the other way around. In the second term, I add until the $$2n + 2$$-th term. I only want to add until $$2n$$, so I have to subtract the $$2n + 2$$-th and $$2n + 1$$-th term. Note that the $$j$$-th term of the series is always $$j$$.

$$\sum_{i=h+1}^{2(h+1)}i=\sum_{i=h+1}^{2h+2}i=\underbrace{\sum_{i=h}^{2h}(i)}_{\text{Hypothesis}}+\underbrace{\sum_{i=1}^{2}}_{(*)}$$

It looks you misunderstand the summation notation. Here are some more details (in expanded forms): $$\sum_{i=h+1}^{2(h+1)}i=\sum_{i=\color{blue}{h+1}}^{\color{red}{2h+2}}i=(\color{blue}{h+1})+(h+2)+\cdots +(2h)+(2h+1)+(\color{red}{2h+2})=\\ \color{green}{h}+(h+1)+(h+2)+\cdots +(2h)+(2h+1)+(2h+2)-\color{green}h=\\ \color{blue}{h}+(h+1)+(h+2)+\cdots +(\color{red}{2h})+(2h+1)+(2h+2)-h=\\ \sum_{i=\color{blue}h}^{\color{red}{2h}}i+(3h+3)=\frac32h(h+1)+3(h+1)=\frac32(h+1)(h+2).$$ Note:

1) In the summations above, the sum is taken from $$\color{blue}{blue}$$ to $$\color{red}{red}$$.

2) In line 2, the number $$\color{green}h$$ is added and subtracted to complete the sum from $$\color{blue}h$$ to $$\color{red}{2h}$$, which is needed for the hypothesis.

You should have said $$\sum_{i=h+1}^{2h+2}i=\left(\sum_{i=h}^{2h}i\right)-h+(2h+1)+(2h+2)$$

• Thanks! I have never descompose a summation wrt the initial values i.e. $i=h+1$. So my questions are: 1) How do you go from $i=h+1$ to $i=h$? 2) Why there is a $-h+(2h+1)+(2h+2)$? – manooooh Jul 8 at 20:47
• In general with integers $a<b<c<d,$ $$\sum_{i=a}^d f(i)=\sum_{i=a}^b f(i) + \sum_{i=b+1}^c f(i) + \sum_{i=c+1}^d f(i);$$ try writing out with some small values of $h$ to see – J. W. Tanner Jul 8 at 20:50

Hint:

Denoting the sums $$S_n$$ and $$S_{n+1}$$, just look at how they begin and end: \begin{alignat}{3} S_n &= {} &n&+{}&(n+1)+(n+2)+\dots &+2n \\ S_{n+1} &={} &&\phantom{+{}}&(n+1)+(n+2)+\dots &+2n +(2n+1)+(2n+2), \end{alignat} so that we have the relation $$S_{n+1}=S_n -n+(2n+1)+(2n+2)=S_n+3(n+1).$$ Can you end the computation?