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The sum of terms, [tex]S_r[/tex], of a series is given by:

[tex]S_r = r^2(r + 1)(r + 2)[/tex]

Given that [tex]u_r[/tex] is the r-th term of the series whose sum is [tex]S_r[/tex], show that:

(a) (i) [tex]u_1 = 6[/tex]

(ii) [tex]u_2 = 42[/tex]

(iii) [tex]u_n = n(n + 1)(4n - 1)[/tex]

Answer :

Final answer:

The values of u₁, u₂, and the formula for un are derived and verified using the given formula for the sum S₁ of the series.


Explanation:

Proof:


To prove the given statements:


  1. (i) u₁ = 6

Substituting r = 1 in the formula for S₁:


S₁ = 1²(1 + 1)(1 + 2) = 6


Since S₁ represents the sum of the first term, u₁, we can conclude that u₁ = 6.


  1. (ii) u₂ = 42

Substituting r = 2 in the formula for S₁:


S₁ = 2²(2 + 1)(2 + 2) = 24


Since S₁ represents the sum of the first two terms, u₁ + u₂, we can calculate u₂ by subtracting u₁ from S₁: u₂ = S₁ - u₁ = 24 - 6 = 18


Therefore, u₂ = 18.


  1. (iii) un = n(n + 1)(4n - 1)

To prove this formula, we can use mathematical induction. First, we can verify that it holds for the base case, n = 1:


u₁ = 1(1 + 1)(4(1) - 1) = 6


Now, assuming the formula holds for some value of n, we can prove that it holds for n + 1:


uₙ₊₁ = (n + 1)((n + 1) + 1)(4(n + 1) - 1)


= (n + 1)(n + 2)(4n + 3)


= (n(n + 1)(4n - 1)) + (6(n + 1)(n + 2)) + 3(n + 1)


= uₙ + 6(n + 1)(n + 2) + 3(n + 1)


= uₙ + (n + 1)(6(n + 2) + 3)


= uₙ + (n + 1)(6n + 15)


By the induction principle, the formula holds for all natural numbers n. Therefore, un = n(n + 1)(4n - 1).


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