Let us assume the required sum = S

Therefore, S = 1² + 2² + 3² + … + n²

Now, we will use the below identity to find the value of S:

n³ - (n – 1)³ = 3n² – 3n + 1

Substituting, n = 1, 2, 3, 4, 5,…, n in the above identity, we get

1³ – (1 – 1)³ = 1³ – 0³ = 3(1)² – 3(1) + 1

2³ – (2 – 1)³ = 2³ – 1³ = 3(2)² – 3(2) + 1

3³ – (3 – 1)³ = 3³ – 2³ = 3(3)² – 3(3) + 1

and so on

n³ - (n – 1)³ = 3n² – 3n + 1

Adding we get,

n³ - 0³ = 3(1² + 2² + 3² + … + n²) - 3(1 + 2 + 3 + 4 + … + n) + (1 + 1 + 1 + 1 + … n times)

Taking common n(n + 1), we get

Thus, the sum of the squares of first n natural numbers =

∴(a) ↔ (iii)

(b) Let us assume the required sum = S

Therefore, S = 1³ + 2³ + 3³ + … + n³

Now, we will use the below identity to find the value of S:

n⁴ - (n – 1)⁴ = 4n³ – 6n² + 4n – 1

Substituting, n = 1, 2, 3, 4, 5,…, n in the above identity, we get

1⁴ – (1 – 1)⁴ = 1⁴ – 0⁴ = 4(1)³ – 6(1)² + 4(1) + 1

2⁴ – (2 – 1)⁴ = 2⁴ – 1⁴ = 4(2)³ – 6(2)² + 4(2) + 1

3⁴ – (3 – 1)⁴ = 3⁴ – 2⁴ = 4(3)³ – 6(3)² + 4(3) + 1

and so on

n⁴ - (n – 1)⁴ = 4n³ – 6n² + 4n – 1

Adding we get,

n⁴ – 0⁴ = 4(1³ + 2³ + 3³ + … + n³) – 6(1² + 2² + 3² + … + n²) +4(1 + 2 + 3 + 4 + … + n) + (1 + 1 + 1 + 1 + … n times)

⇒ n⁴ = 4S – n(n + 1)(2n + 1) + 2n (n + 1) – n

⇒ 4S = n⁴ + n + n(n + 1)(2n + 1) – 2n(n + 1)

⇒ 4S = n(n³ + 1) + n(n + 1)[(2n + 1) – 2]

⇒ 4S = n[n³ + 1 + (n + 1)(2n – 1)]

⇒ 4S = n[n³ + 1 + 2n² – n + 2n – 1]

⇒ 4S = n[n³ + 2n² + n]

⇒ 4S = n²(n² + 2n + 1)

⇒ 4S = n²(n + 1)²

Thus, the sum of the cubes of first n natural numbers =

∴(b) ↔ (i)

(c) Let S_n = 2 + 4 + 6 + … + 2n

= 2(1 + 2 + 3 + … + n)

= n(n + 1)

∴ (a) ↔ (ii)

(d) Let S be the required sum.

Therefore, S = 1 + 2 + 3 + 4 + 5 + … + n

Clearly, it is an Arithmetic Progression whose first term = 1, last term = n and number of terms = n.

Using the formula,

Therefore,

or we can say that,

∴ (d) ↔ (iv)