Q.Match the entries in Column I with Column II. Column I:

The key idea is to recall the standard formulas for the sum of arithmetic series and sums of powers of natural numbers.

1. The sum of the squares of the first $n$ natural numbers, $1^2 + 2^2 + \ldots + n^2$, is given by $\dfrac{n(n+1)(2n+1)}{6}$. This matches (a) with (iii).
2. The sum of the cubes of the first $n$ natural numbers, $1^3 + 2^3 + \ldots + n^3$, is given by $\left(\dfrac{n(n+1)}{2}\right)^2$. This matches (b) with (i). …

Understanding how to sum common series is a fundamental skill in mathematics, especially for competitive exams. These formulas often appear in various contexts, from basic algebra to calculus and probability. The key is to recognise the pattern of the terms in the series and apply the appropriate formula.

Let's go through each series in Column I and match it with its sum in Column II.

1. Series (a): $1^2 + 2^2 + 3^2 + \ldots + n^2$

   This is the sum of the squares of the first $n$ natural numbers. This is a standard series whose sum is given by a well-known formula.

   The sum of the squares of the first $n$ natural numbers is:

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

   Comparing this with Column II, we see that it matches entry (iii).

   Therefore, (a) $\leftrightarrow$ (iii).

2. Series (b): $1^3 + 2^3 + 3^3 + \ldots + n^3$

   This is the sum of the cubes of the first $n$ natural numbers. This is another standard series, and its sum has a particularly elegant relationship with the sum of the first $n$ natural numbers.

   The sum of the cubes of the first $n$ natural numbers is:

   $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$

   Comparing this with Column II, we find that it matches entry (i).

   Therefore, (b) $\leftrightarrow$ (i).

3. Series (c): $2 + 4 + 6 + \ldots + 2n$

   This series consists of the first $n$ even natural numbers. We can factor out a 2 from each term to simplify it.

   $2 + 4 + 6 + \ldots + 2n = 2(1 + 2 + 3 + \ldots + n)$

   Now, the expression inside the parenthesis is the sum of the first $n$ natural numbers, which is a very common arithmetic progression.

   The sum of the first $n$ natural numbers is:

   $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ …