Exercise 4.3.9 ($\sum_{x=1}^n x = n(n+1)/2$)

Question

By use of the algebraic identity $(x+1)^2 - x^2 = 2x + 1$, establish that $(n+1)^2 - 1^2 = \sum_{x=1}^n{(x+1)^2 -x^2} = \sum_{x=1}^n (2x+1)$ and so derive the result $\sum_{x=1}^n x = n(n+1)/2$.

Richard Ganaye · Accepted Answer

\begin{proof} 
The usual properties of sums give
\begin{align*}
\sum_{i=1}^n [(i+1)^2 - i^2]&= \sum_{i=1}^n (i+1)^2  - \sum_{i=1}^n i^2 \
&=\sum_{j = 2}^{n+1} j^2 - \sum_{j=1}^n j^2&(j=i +1 	ext{ in the first sum})\
&=\left((n+1)^2 + \sum_{j = 2}^{n} j^2 ight) - \left( 1 + \sum_{j=2}^n j^2 ight)\
&=(n+1)^2 - 1.
\end{align*}
Since $(i+1)^2 - i^2 = 2i + 1$ for all $i$, we obtain
\begin{align}
\sum_{i=1}^n (2 i + 1) = (n+1)^2 - 1.
\end{align}
(So the sum of the $n$ first odd integers is equal to $n^2$.)

We define $S(n) = \sum_{i=1}^n i$ for all positive integer $n$. By (1),
\begin{align*}
(n+1)^2 - 1 &= \sum_{i=1}^n (2 i + 1)\
&= 2 \sum_{i=1}^n i + \sum_{i=1}^n 1\
&=2 S(n) + n.
\end{align*}
Therefore
\begin{align*}
S(n) &= \frac{(n+1)^2 - (n+1)}{2}\
&=\frac{n(n+1)}{2}.
\end{align*}
For all positive integer $n$,
$$\sum_{i=1}^n i = \frac{n(n+1)}{2}.$$
\end{proof}

Exercise 4.3.9 ($\sum_{x=1}^n x = n(n+1)/2$)

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Exercise 4.3.9 ($\sum_{x=1}^n x = n(n+1)/2$)

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