Sequences & Series
Sequences And Series
nta_abhyas_2025
Grade 11
Question:
If the sum $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \ldots$ up to 20 terms is equal to $\frac{n}{21}$, then $n$ is equal to
Step-by-Step Solution
Key Concept: Recognize that the denominator uses the sum of squares formula and apply partial fraction decomposition to create a telescoping series
The sum of squares formula gives $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$. The general term is $\frac{1}{\frac{n(n+1)(2n+1)}{6}} = \frac{6}{n(n+1)(2n+1)}$. Using partial fractions: $\frac{6}{n(n+1)(2n+1)} = 6\left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(2n+1)}\right)$. Summing telescopically from $n=1$ to $20$: $S_{20} = 6\left[\frac{1}{1 \cdot 2} - \frac{1}{21 \cdot 41}\right] = 6\left(1 - \frac{1}{21}\right) = \frac{6 \cdot 20}{21}$.
Correct Answer: 6.20/21