Question

#45points


Generalise the following statement


1 + 4 + 9 + 16 .... + n^2 ​

Answer 1

Step-by-step explanation:

Given Series is 1² + 2² + 3² + 4² + .... + n²

First, remember that (a - b)³ = a³ - b³ - 3a²b + 3ab², now lets apply in (k - 1).

Consider the identity:

k³ - (k - 1)³ = k³ - (k³ - 1 - 3k² + 3k)

                 = k³ - k³ + 1 + 3k² - 3k

                 = 3k² - 3k + 1

Let us put k = 1,2,3...n, we get

(i)  1³ - (1 - 1)³ = 3(1)² - 3(1) + 1

⇒ 1³ - 0³ = 3(1)² - 3(1) + 1

(ii) 2³ - (2 - 1)³ = 3(2)² - 3(2) + 1

⇒ 2³ - (1)³ = 3(2)² - 3(2) + 1

(iii) 3³ - (3 - 1)³ = 3(3)² - 3(3) + 1

⇒ 3³ - (2)³ = 3(3)² - 3(3) + 1

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⇒ n³ - (n - 1)³ = 3(n)² - 3(n) + 1

On adding all the equations, we get

⇒ n³ - 0³ = 3(1² + 2² + 3² + 4² + ..... + n²) - 3(1 + 2 + 3 + 4 + ... +n) + n

⇒ n³ = 3(S) - 3 * [n(n + 1)/2] + n

⇒ 3S =  n³ + (3/2)n(n + 1) - n

⇒ 3S = n³ - n + [(3/2)n(n + 1)]

⇒ 3S = n(n² - 1) + [(3/2)n(n + 1)]

⇒ 3S = n(n + 1)(n - 1 + 3/2)

⇒ 3S = n(n + 1)[2n - 2 + 3/2]

⇒ 3S = n(n + 1)(2n + 1)/2

⇒ S = n(n + 1)(2n + 1)/6

Hope it helps!

Answer 2

Step-by-step explanation:

Let us assume required be K.

Therefore, K = 1² + 2² + 3² + 4² + .... + n²

Consider the identity,

n³ - (n - 1)³ = 3n² - 3n + 1

Substitute n = 1,2,...,n we get

1³ - 0³ = 3 * 1² - 3 * 1 + 1

2³ - 1³ = 3 * 2² - 3 * 2 + 1

.

.

n³ - (n - 1)³ = 3 * n² - 3 * n + 1

Adding both the sides of the equation

n³ - 0³ = 3(1² + 2² + 3² + 4² + .. + n²) - 3(1 + 2 + 3 + 4 + ... + n) + (1 + 1 + 1 + ... n)

n³ = 3K - 3[n(n+1)/2] + n

n³ = 3K - (3/2)k(k+1) - n

3K = n(n² - 1) + (3/2)n[n + 1]

3K = n(n + 1)(n - 1 + 3/2)

3K = n(n+1)(2n-2+3/2)

3K = n(n+1)(2n+1)/2

K = n(n+1)(2n+1)/6

Hope it helps you