Question

Let {a_n}_n=1^∞ be a sequence such that a₁ = 1, a₂ = 1 and a_{n + 2} = 2a_{n + 1} + a_n for all n ≥1. Then the value of \(47\displaystyle\sum^\infty_{n=1}\frac{a_n}{2^{3n}}\) is equal to ________.

47∑^∞_n=1 a_n/2³ⁿ

Answer 1

a_n+2 = 2a_n+1 + a_n has its characteristic equation as

x² = 2x + 1

⇒ x = 1 ± √2

So, a_n = a(1 + √2)ⁿ⁻¹ + b(1 − √2)ⁿ⁻¹

∵ a₁ = 1

⇒ a + b = 1

and a₂ = 1

⇒ (a + b) + √2(a − b) = 1

⇒ a = \(\frac 12\) and \(b = \frac 12\)

So, a_n = \(\frac {(1 + \sqrt 2)^{n - 1}+ (1 - \sqrt 2)^{n - 1}}{2}\)

\(\sum \limits^{\infty}_{n = 1} \frac{a_n}{2^{3n}} = \frac 1{16} \left[ \sum\limits_{n = 1}^\infty \left(\frac{1 + \sqrt 2}8\right)^{n - 1} + \sum\limits_{n = 1}^\infty \left(\frac{1 - \sqrt 2}8\right)^{n - 1}\right]\)

\(=\frac 1{16} \left[ \frac 8{7 - \sqrt 2} + \frac 8{7 + \sqrt 2}\right]\)

\(= \frac 7{47}\)

\(\therefore 47 \sum \limits_{n= 1}^\infty \frac{a_n}{2^{3n}} = 7\)

Answer 2

Answer is: 7

Divide by 8ⁿ we get

64P – 8 – 1 = 16 P – 2 + P

47P = 7