The nth term of $\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+..........$ is equal to :

Question

Accepted Answer

To determine why Option 2, $\frac{2n+1}{3}$, is the correct answer for the nth term of the series $\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots$, we need to analyze the structure of the series.

### Explanation for Option 2:

The nth term of the series is given by:

$$
T_n = \frac{1^2 + 2^2 + 3^2 + \ldots + n^2}{1 + 2 + 3 + \ldots + n}
$$

#### Numerator Analysis:
The numerator is the sum of squares of the first n natural numbers. The formula for the sum of squares is:

$$
1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}
$$

#### Denominator Analysis:
The denominator is the sum of the first n natural numbers. The formula for this sum is:

$$
1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}
$$

#### Simplifying the nth Term:
Now, substituting these formulas into the expression for $T_n$:

$$
T_n = \frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}
$$

Simplifying this expression:

$$
T_n = \frac{n(n+1)(2n+1)}{6} \times \frac{2}{n(n+1)}
$$

Canceling $n(n+1)$ from the numerator and the denominator:

$$
T_n = \frac{2(2n+1)}{6} = \frac{2n+1}{3}
$$

This confirms that the nth term of the series is indeed $\frac{2n+1}{3}$, which matches Option 2.

### Explanation for Why Other Options are Incorrect:

1. **Option 1: $\frac{2n+1}{6}$**
   - This option suggests a different simplification of the series. However, as shown in the simplification process, the correct factor after canceling terms is $\frac{2n+1}{3}$, not $\frac{2n+1}{6}$.

2. **Option 3: $\frac{n(2n+1)}{2}$**
   - This option does not account for the division by the sum of the first n natural numbers in the denominator. The correct simplification involves dividing by $\frac{n(n+1)}{2}$, leading to $\frac{2n+1}{3}$.

3. **Option 4: $\frac{2n-1}{3}$**
   - This option incorrectly alters the numerator term. The correct numerator after simplification is $2n+1$, not $2n-1$.

In conclusion, the correct answer is Option 2, $\frac{2n+1}{3}$, as it accurately represents the nth term of the given series after proper simplification.