Chapter F, Problem 36E

To determine

To find: The number n such that ${\displaystyle \sum _{i=1}^{n}i}=78$.

Answer to Problem 36E

The number n is 12.

Explanation of Solution

Theorem used:

Let c be a constant and n be a positive integer. Then,

${\displaystyle \sum _{i=1}^{n}c}=nc$, ${\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{2}$ and ${\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$.

Calculation:

By the above theorem, it is given that ${\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{2}$.

Thus, the expression ${\displaystyle \sum _{i=1}^{n}i}=78$ simplified as, $\frac{n\left(n+1\right)}{2}=78$.

Solve the equation $\frac{n\left(n+1\right)}{2}=78$ and obtain the value of n.

$\begin{array}{l}\frac{n\left(n+1\right)}{2}=78\\ n\left(n+1\right)=156\\ n^2+n-156=0\end{array}$

The equation $n^2+n-156=0$ is a quadratic equation with variable n.

Here, the value of $a=1,b=1andc=-156$.

Solve the quadratic equation $n^2+n-156=0$.

$\begin{array}{c}n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(-156\right)}}{2\left(1\right)}\\ =\frac{-1\pm \sqrt{625}}{2}\\ =\frac{-1\pm 25}{2}\\ =12\text{or}-13\end{array}$

That is, the number n is $12\text{or}-13$.

The sum ${\displaystyle \sum _{i=1}^{n}i}=78$ is positive thus, the number $n=-13$ is not possible.

Therefore, the sum ${\displaystyle \sum _{i=1}^{n}i}=78$ as when $n=12$.