Answer
$4901$
Work Step by Step
We have to determine the sum:
$S=8+8\dfrac{1}{4}+8\dfrac{1}{2}+8\dfrac{3}{4}+9+...+50$
As $8\dfrac{1}{4}-8=8\dfrac{1}{2}-8\dfrac{1}{4}=8\dfrac{3}{4}-\dfrac{1}{2}=...=\dfrac{1}{4}$, the sequence is arithmetic.
Determine its first term and the common difference:
$a_1=8$
$d=\dfrac{1}{4}$
Determine the number of terms:
$a_n=a_1+(n-1)d$
$a_n-a_1=(n-1)d$
$n-1=\dfrac{a_n-a_1}{d}$
$n=\dfrac{a_n-a_1}{d}+1$
$n=\dfrac{50-8}{\dfrac{1}{4}}+1$
$n=169$
Therefore, the given sum contains $169$ terms, so we have to determine the sum of the first $169$ terms. We use the formula:
$S_n=\dfrac{n(a_1+a_n)}{2}$
$8+8\dfrac{1}{4}+8\dfrac{1}{2}+8\dfrac{3}{4}+9+...+50=\dfrac{169(8+50)}{2}$
$=\dfrac{169\cdot 58}{2}$
$=4901$
