Question

Sum both sides of the identity${\mathrm{k}}^2-(\mathrm{k}-1{)}^2=2\mathrm{k}-1\text{from}\mathrm{k}=1\text{to}\mathrm{k}=\mathrm{n}$ to and use Exercise 35 to find

a) a formula for$\sum _{k=1}^n (2k-1)$ (the sum of the first n odd natural numbers).

b) a formula for $\sum _{k=1}^n \mathrm{k}$

Short answer

Thus, the formula for the following is given below:

$\begin{array}{r}\sum _{\mathrm{k}=1}^{\mathrm{n}} (2\mathrm{k}-1)={\mathrm{n}}^2\\ \sum _{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}=\frac{{\mathrm{n}}^2+\mathrm{n}}{2}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\end{array}$

Step-by-step solution

Step 1:

Given:${\mathrm{k}}^2-(\mathrm{k}-1{)}^2=2\mathrm{k}-1$

1. If the given equality holds, then the sum each side over the same values of also has to be equal:

$\sum _{k=1}^n (2k-1)=\sum _{k=1}^n \left(k^2-(k-1{)}^2\right)$

Let $a_k=k^2$

$=\sum _{k=1}^n \left(a_k-a_{k-1}\right)$

Result of the previous exercise (35) $\sum _{j=1}^n \left(a_j-a_{j-1}\right)=a_n-a_0$

$\begin{array}{r}=a_n-a_0\\ =n^2-0^2\\ =n^2\end{array}$

Thus, we obtain$\sum _{k=1}^n (2k-1)=n^2$

Step 2:

1. From the result (a)$\sum _{k=1}^n (2k-1)=n^2$

The sum of a difference is the difference of the sums over each term.

$\sum _{k=1}^n \left(2k\right)-\sum _{k=1}^n 1=n^2$

We can factor out 2 in the first sum and $\sum _{\mathrm{k}=1}^{\mathrm{n}} 1=\mathrm{n}\text{:}$

$2\sum _{k=1}^n \mathrm{k}-\mathrm{n}={\mathrm{n}}^2$

Add to each side of the previous equation:

$2\sum _{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}={\mathrm{n}}^2+\mathrm{n}$

Divide each side of the previous equation by :

$\sum _{k=1}^n k=\frac{n^2+n}{2}=\frac{n(n+1)}{2}$