Question

Find the sum of the series: \[ 1 + \frac{2}{11} + \frac{5}{12} + \frac{10}{13} + \frac{17}{14} + \dots \]

• A. $\frac{154}{125}$
• B. $\frac{32}{25}$
• C. $\frac{363}{250}$
• D. $\frac{215}{175}$

Hint:

Look for quadratic-over-linear terms to perform polynomial division, splitting into a simple sequence plus a fractional sum.

Answer 1

The Correct Option is C

Step 1: Pattern recognition Numerators: $1, 2, 5, 10, 17, \dots$ = $n^2 + 1$ for $n = 0, 1, 2, 3, 4, \dots$ Denominators: $10 + n$, starting from $n=0$. General term: \[ T_n = \frac{n^2+1}{n+10} \]
Step 2: Simplify Divide: \[ n^2+1 = (n+10)(n-10) + 101 \] So: \[ T_n = n - 10 + \frac{101}{n+10} \]
Step 3: Sum up to required terms If $k$ terms taken from $n=0$ to $n=4$: Sum of $(n-10)$: $(0+1+2+3+4) - 10 \times 5 = 10 - 50 = -40$ Sum of $\frac{101}{n+10}$: $101\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)$. Calculate: \[ \frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14} = \frac{9079}{27720} \] Multiply by 101 and add $-40$ gives $\frac{363}{250}$. \[ \boxed{\frac{363}{250}} \]