Question

Let ${\displaystyle {\left\{a_n\right\}}_{n=0}^{\infty }}$ be a sequence such that a₀ = a₁ = 0 and a_n+2 = 2a_n+1 – a_n + 1 for all n ≥ 0. Then, ${\displaystyle \sum _{n=2}^{\infty }\frac{a^n}{7^n}}$ is equal to ______.

विकल्प

• ${\displaystyle \frac{6}{343}}$

• ${\displaystyle \frac{7}{216}}$

• ${\displaystyle \frac{8}{343}}$

• ${\displaystyle \frac{49}{216}}$

Answer 1

Let ${\displaystyle {\left\{a_n\right\}}_{n=0}^{\infty }}$ be a sequence such that a₀ = a₁ = 0 and a_n+2 = 2a_n+1 – a_n + 1 for all n ≥ 0. Then, ${\displaystyle \sum _{n=2}^{\infty }\frac{a^n}{7^n}}$ is equal to ${\displaystyle \underline{\frac{7}{216}}}$.

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 = ${\displaystyle \frac{n(n-1)}{2}}$

Let p = ${\displaystyle \sum _{n=2}^{\infty }\frac{a^n}{7^n}}$

⇒ p = ${\displaystyle \sum _{n=2}^{\infty }\frac{n(n-1)}{{2.7}^n}}$

⇒ p = ${\displaystyle \frac{1}{7^2}+\frac{3}{7^3}+\frac{6}{7^4}+\frac{10}{7^5}+......}$ ....(i)

⇒ ${\displaystyle \frac{p}{7}=\frac{1}{7^3}+\frac{3}{7^4}+\frac{6}{7^5}+\frac{10}{7^6}+......}$  ....(ii)

Equation (i) – equation (ii), we get

${\displaystyle \frac{6p}{7}=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+\frac{4}{7^5}+......}$ ....(iii)

⇒ ${\displaystyle \frac{6p}{7^2}=\frac{1}{7^3}+\frac{2}{7^4}+\frac{3}{7^5}+\frac{4}{7^6}+......}$ ...(iv)

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

${\displaystyle \frac{6p}{7}.(1-\frac{1}{7})=\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+\frac{1}{7^5}+......}$

⇒ ${\displaystyle \frac{6p}{7}\left(\frac{6}{7}\right)=\frac{\frac{1}{7^2}}{1-\frac{1}{7}}}$

⇒ ${\displaystyle {\left(\frac{6}{7}\right)}^{2}p=\frac{1}{\left(7\right)\left(6\right)}}$

⇒ p = ${\displaystyle \frac{7}{216}}$