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

Question

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

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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!!

Answer 1

\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.

Answer 2

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 $j$.

Answer 3

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

Answer 4

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?

Answer 5

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

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.