The sequence \((x_n)\), where \(n \geq 0\), is defined by \(x_0 =1\) and

\[x_n = \sum_{k=0}^{n-1}x_k \quad \text{for} \quad n \geq 1.\]

The sum

\[\sum_{k=0}^\infty \dfrac{1}{x_k}\]

equals

- \(1\), (b) \(\dfrac{6}{5}\), (c) \(\dfrac{8}{5}\), (d) \(3\), (e) \(\dfrac{27}{5}\).

Can we write the sequence \((x_n)\) out?

Using the rule, we have:

\(x_0 = 1\),

\(x_1 =x_0= 1\),

\(x_2=x_0+x_1=1+1=2\),

\(x_3 =x_0+x_1+x_2= 1+1+2 = 4\),

\(x_4=x_0+x_1+x_2+x_3 = 1+1+2+4 = 8, ...\)

So \(x_n = 2^{n-1}\) for \(n \geq 1\), and apart from the first term, we have a geometric sequence.

We also have that \(\dfrac{1}{x_n} = \dfrac{1}{2^{n-1}}\), and so the sequence \(\left(\dfrac{1}{x_n}\right)\) is also geometric except for the first term.

We can remember now that the sum to infinity of a geometric progression with first term \(a\) and common ratio \(r\) is \(\dfrac{a}{1-r}\).

So \[\sum_{k=0}^\infty \dfrac{1}{x_k} = 1 + 1 + \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...\]

\[= 1 + \dfrac{1}{1-\dfrac{1}{2}} = 3,\]

and so that answer is (d).