Question

For the sequence aᵢ=i(i+1) and the partial sum of its first n terms Sₙ=∑ᵢ=1ⁿ aᵢ, find

(a) a₄
(b) S₄
(c) S ₅₀-S₄₉

Answer 1

a) a₄ = 20, b) S₄ = 39, c) S₅₀-S₄₉ = 2550

Step-by-step explanation:

To find a₄, we substitute i with 4 in the formula for aᵢ. Therefore, a₄ = 4(4+1) = 20.

To find S₄, we substitute n with 4 in the formula for Sₙ. Therefore, S₄ = ∑ᵢ=1⁴ aᵢ = a₁ + a₂ + a₃ + a₄ = 1(1+1) + 2(2+1) + 3(3+1) + 4(4+1) = 1 + 6 + 12 + 20 = 39.

To find S₅₀-S₄₉, we substitute n with 50 and 49 in the formula for Sₙ. Therefore, S₅₀-S₄₉ = ∑ᵢ=₁⁵₀ aᵢ - ∑ᵢ=₁⁴₉ aᵢ = (a₁ + a₂ + ... + a₅₀) - (a₁ + a₂ + ... + a₄₉). Since there are common terms between the two sums, we can cancel them out, leaving only a₅₀. Therefore, S₅₀-S₄₉ = a₅₀ = 50(50+1) = 2550.