## Calculus: Early Transcendentals 8th Edition

$12$
Find the value of $n$ for the sum $\sum\limits_{i =1}^{n}i=78$ $\sum\limits_{i =1}^{n}i=\frac{n(n+1)}{2}$ Thus, $\frac{n(n+1)}{2}=78$ $n^{2}+n=78\times2$ $n^{2}+n=156$ $n^{2}+n-156=0$ $(n+13)(n-12)=0$ $n=-13, 12$ Through neglecting the negative value of $n$, we have $n=12$ Hence, $n=12$