#45points
Generalise the following statement
1 + 4 + 9 + 16 .... + n^2
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#45points
Generalise the following statement
1 + 4 + 9 + 16 .... + n^2
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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!