Question

Let \(t_n=\frac{1}{n}\displaystyle\sum^{n}_{k=1}\left(\frac{k}{n} \right)^2\)for n=1,2,3…. Then t₁₀ is equal to

• A. \(\frac{7}{600}\)
• B. \(\frac{231}{100}\)
• C. \(\frac{209}{600}\)
• D. \(\frac{11}{200}\)
• E. \(\frac{77}{200}\)

Answer 1

The Correct Option is B

We are given that: \[ t_n = \frac{1}{n} \sum_{k=1}^{n} \left( \frac{k}{n} \right)^2 \] This can be written as: \[ t_n = \frac{1}{n} \sum_{k=1}^{n} \frac{k^2}{n^2} = \frac{1}{n^3} \sum_{k=1}^{n} k^2 \] Using the formula for the sum of squares: \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \] Substitute this into the expression for \( t_n \): \[ t_n = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2} \] Now, we need to calculate \( t_{10} \). For \( n = 10 \): \[ t_{10} = \frac{(10+1)(2 \cdot 10 + 1)}{6 \cdot 10^2} = \frac{11 \cdot 21}{6 \cdot 100} = \frac{231}{600} = \frac{231}{100} \] 

The correct option is (B) : \(\frac{231}{100}\)

Answer 2

The given expression for \( t_n \) is:

\[ t_n = \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^2 \]

We need to find \( t_{10} \), so substitute \( n = 10 \) into the formula:

\[ t_{10} = \frac{1}{10} \sum_{k=1}^{10} \left(\frac{k}{10}\right)^2 \] \[ t_{10} = \frac{1}{10} \sum_{k=1}^{10} \frac{k^2}{100} = \frac{1}{10} \times \frac{1}{100} \sum_{k=1}^{10} k^2 \]

Now, we need to calculate \( \sum_{k=1}^{10} k^2 \). The formula for the sum of squares of the first \(n\) natural numbers is:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

Substituting \( n = 10 \) into this formula:

\[ \sum_{k=1}^{10} k^2 = \frac{10(10+1)(2 \times 10 + 1)}{6} = \frac{10 \times 11 \times 21}{6} = \frac{2310}{6} = 385 \]

Now, substitute this value back into the expression for \( t_{10} \):

 \[ t_{10} = \frac{(10+1)(2 \cdot 10 + 1)}{6 \cdot 10^2} = \frac{11 \cdot 21}{6 \cdot 100} = \frac{231}{600} = \frac{231}{100} \] 

Thus, \( t_{10} \) is \( \frac{231}{100}\).