Question

nth Term of a Sequence Find a formula for the nth term of the sequence.

$\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}},\mathrm{.....}$

[Hint: Write each term as a power of 2.]

Short answer

The formula for the nth term of the sequence is $2^{\left(1-\frac{1}{2^n}\right)}$.

Step-by-step solution

Step 1.  Given

The given sequence is$\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}},\mathrm{.....}$

Step 2.  To determine

We have to write aformula for the nth term of the given sequence.

Step 3.  Calculation

We try to write the given sequence using a general term with n.

First term is = $\sqrt{2}=2^{\frac{1}{2}}$.

Second term is = $\sqrt{2\sqrt{2}}=\sqrt{2\times 2^{\frac{1}{2}}}=\sqrt{2^{\frac{3}{2}}}={\left(2^{\frac{3}{2}}\right)}^{\frac{1}{2}}=2^{\frac{3}{2}\times \frac{1}{2}}=2^{\frac{3}{4}}$.

Third term is = $\sqrt{2\sqrt{2\sqrt{2}}}=\sqrt{2\left(\sqrt{2\sqrt{2}}\right)}=\sqrt{2\left(2^{\frac{3}{4}}\right)}=\sqrt{2^{1+\frac{3}{4}}}=\sqrt{2^{\frac{7}{4}}}={\left(2^{\frac{7}{4}}\right)}^{\frac{1}{2}}=2^{\frac{7}{8}}$.

Fourth term is = $\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}=\sqrt{2\left(\sqrt{2\sqrt{2\sqrt{2}}}\right)}=\sqrt{2\left(2^{\frac{7}{8}}\right)}=\sqrt{2^{\frac{15}{8}}}={\left(2^{\frac{15}{8}}\right)}^{\frac{1}{2}}=2^{\frac{15}{16}}$.

So, the exponents of 2 are: $\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16},....$

We’ll try to write them using a common pattern.

$\begin{array}{l}\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16},\mathrm{....}\\ =\frac{2-1}{2},\frac{4-1}{4},\frac{8-1}{8},\frac{16-1}{16},\mathrm{....}\\ =\left(\frac{2}{2}-\frac{1}{2}\right),\left(\frac{4}{4}-\frac{1}{4}\right),\left(\frac{8}{8}-\frac{1}{8}\right),\left(\frac{16}{16}-\frac{1}{16}\right),\mathrm{....}\\ =\left(1-\frac{1}{2}\right),\left(1-\frac{1}{4}\right),\left(1-\frac{1}{8}\right),\left(1-\frac{1}{16}\right),\mathrm{....}\\ =\left(1-\frac{1}{2^1}\right),\left(1-\frac{1}{2^2}\right),\left(1-\frac{1}{2^3}\right),\left(1-\frac{1}{2^4}\right),\mathrm{....}\end{array}$

So, we see a pattern that the nth exponent is $\left(1-\frac{1}{2^n}\right)$.

Hence, the nth term is given by: $2^{\left(1-\frac{1}{2^n}\right)}$.