Question

Let `{a_n}_(n = 0)^∞` be a sequence such that a₀ = a₁ = 0 and a_n+2 = 2a_n+1 – a_n + 1 for all n ≥ 0. Then, `sum_(n = 2)^∞ a^n/7^n` is equal to ______.

Options

• `6/343`

• `7/216`

• `8/343`

• `49/216`

Answer 1

Let `{a_n}_(n = 0)^∞` be a sequence such that a₀ = a₁ = 0 and a_n+2 = 2a_n+1 – a_n + 1 for all n ≥ 0. Then, `sum_(n = 2)^∞ a^n/7^n` is equal to `underlinebb(7/216)`.

Explanation:

Given, a₀ = a₁ = 0 and a_n+2 = 2a_n+1 – a_n + 1 ∀ n ≥ 0

⇒ a₂ = 2a₀ – a₀ + 1 = 1

and a₃ = 2a₂ – a₁ + 1 = 3

and a₄ = 2a₃ – a₂ + 1 = 6

And a₅ = 2a₄ – a₃ + 1 = 10

∴ a_n = `(n(n - 1))/2`

Let p = `sum_(n = 2)^∞ a^n/7^n`

⇒ p = `sum_(n = 2)^∞ (n(n - 1))/2.7^n`

⇒ p = `1/7^2 + 3/7^3 + 6/7^4 + 10/7^5 + ......` ....(i)

⇒ `p/7 = 1/7^3 + 3/7^4 + 6/7^5 + 10/7^6 + ......`  ....(ii)

Equation (i) – equation (ii), we get

`(6p)/7 = 1/7^2 + 2/7^3 + 3/7^4 + 4/7^5 + ......` ....(iii)

⇒ `(6p)/7^2 = 1/7^3 + 2/7^4 + 3/7^5 + 4/7^6 + ......` ...(iv)

Equation (iii) – equation (iv), we get

`(6p)/7.(1 - 1/7) = 1/7^2 + 1/7^3 + 1/7^4 + 1/7^5 + ......`

⇒ `(6p)/7(6/7) = (1/7^2)/(1 - 1/7)`

⇒ `(6/7)^2p = 1/((7)(6))`

⇒ p = `7/216`