Question

Column I	Column II
(a) 12 + 22 + 32 + ...+ n2	(i) `((n(n + 1))/2)^2`
(b) 13 + 23 + 33 + ... + n3	(ii) n(n + 1)
(c) 2 + 4 + 6 + ... + 2n	(iii) `(n(n + 1)(2n + 1))/6`
(d) 1 + 2 + 3 +...+ n	(iv) `(n(n + 1))/2`

Answer 1

Column I	Column II
(a) 12 + 22 + 32 + ...+ n2	(i) `(n(n + 1)(2n + 1))/6`
(b) 13 + 23 + 33 + ... + n3	(ii) `((n(n + 1))/2)^2`
(c) 2 + 4 + 6 + ... + 2n	(iii) n(n + 1)
(d) 1 + 2 + 3 +...+ n	(iv) `(n(n + 1))/2`

Explanation:

(a) Let S_n = 1² + 2² + 3² + … + n2

K³ – (K – 1)³ = 3K² – 3K + 1

For K = 1, 1³ – 0³ = 3(1)² – 3(1) + 1

For K = 2, 2³ – 1³ = 3(2)² – 3(2) + 1

For K = 3, 3³ – 2³ = 3(3)² – 3(3) + 1

… … …

For K = n, n³ – (n – 1)³ = 3(n)² – 3(n) + 1

Adding Column wise, we get

n³ – 0³ = 3(1² + 2² + 3² + … + n²) – 3(1 + 2 + 3 + … + n) + n

⇒ n³ = `3 * "S"_n - (3n(n + 1))/2 + n`

⇒ `n^3 + (3n(n + 1))/2 - n = 3 * "S"_n`

⇒ `(2n^3 + 3n^2 + 3n - 2n)/2 = 3 * "S"_n`

⇒ 6 · S_n = 2n³ + 3n² + n

⇒ 6 · S_n = n(2n² + 3n + 1)

⇒ 6 · S_n = n[2n² + 2n + n + 1]

⇒ 6 · S_n = n(n + 1)(2n + 1)

⇒ `"S"_"n" = ("n"("n" + 1)(2"n" + 1))/6`

(b) Let S_n = 1³ + 2³ + 3³ + … + n³

K⁴ – (K – 1)⁴ = 4K³ – 6K² + 4K – 1

For K = 1

1⁴ – 0⁴ = 4(1)⁴ – 6(1)² + 4(1) – 1

For K = 2

2⁴ – 1⁴ = 4(2)³ –6(2)² + 4(2) – 1

For K = 3

3⁴ – 2⁴ = 4(3)³ – 6(3)² + 4(3) – 1

… … …

For K = n

n⁴ – (n –1)⁴ = 4(n)³ – 6(n²) + 4(n) – 1

Adding column wise, we get

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

⇒ n⁴ = `4 * "S"_n = (6n(n + 1)(2n + 1))/6 + (4n(n + 1))/2 - n`

⇒ n⁴ = `4 * "S"_n - n(n + 1)(2n + 1) + 2n(n + 1) - n`

⇒ `n^4 + n(n + 1)(2n + 1) - 2n(n + 1) + n = 4 * "S"_n`

⇒ `n[n^3 + (n + 1)(2n + 1) - 2n - 2 + 1] = 4 * "S"_n`

⇒ `n[n^3 + 2n^2 + 3n + 1 - 2n - 1] = 4 * "S"_n`

⇒ `n[n^3 + 2n^2 + n] = 4 * "S"_n`

⇒ `(n^2(n^2 + 2n + 1))/4 = "S"_n`

⇒ `(n^2(n + 1)^2)/4 = "S"_n`

∴ `S_n = [(n(n + 1))/2]^2`

(c) Let S_n = 2 + 4 + 6 + .... + 2n

= 2(1 + 2 + 3 + .... + n)

= `2 (n(n + 1))/2`

= n(n + 1)

(d) Let S_n = 1 + 2 + 3 + … + n

= `(n(n + 1))/2`