Question

How do you find the ${{n}^{th}}$ term of the sequence $1,1\dfrac{1}{2},1\dfrac{3}{4},1\dfrac{7}{8},...$?

Answer 1

Hint: Now let us first consider the given sequence. In the given sequence we have a mixed fraction. We will first convert them into normal fractions by using $a\dfrac{p}{q}=\dfrac{aq+p}{q}$. Now we will try to find the relation between numerator and denominator for ${{n}^{th}}$ term and hence write the general term of the sequence.

Complete step-by-step solution:
Consider the given sequence $1,1\dfrac{1}{2},1\dfrac{3}{4},1\dfrac{7}{8},...$
Now to solve the equation let us simplify the mixed fraction first.
Now we know that $a\dfrac{p}{q}=\dfrac{aq+p}{q}$ Hence using this we will convert the mixed fraction into normal fraction.
Hence we get the sequence as $1,\dfrac{2\times 1+1}{2},\dfrac{3\times 1+4}{4},\dfrac{8\times 1+7}{8},...$
$\Rightarrow 1,\dfrac{3}{2},\dfrac{7}{4},\dfrac{15}{8},...$
Now let us find the pattern in the above equation.
Consider the first term. Numerator of the first term is 1 and the denominator is also 1. Here we can write the numerator as 2 × 1 – 1 which is nothing but 2 × denominator – 1.
Now consider the second term. Numerator of the second term is 3 and the denominator is 2. Here too we can write the numerator as 2 × 2 – 1. Hence numerator = 2 × denominator – 1.
Now again consider the third term. Here too we can see that numerator = 15 = 2 × denominator – 1.
Now even if we check the denominator the denominator of ${{n}^{th}}$ term is given by ${{2}^{n-1}}$ .
Hence we can say that ${{n}^{th}}$ term of the sequence is given by $\dfrac{{{2}^{n-1}}\times 2-1}{{{2}^{n-1}}}=\dfrac{{{2}^{n}}-1}{{{2}^{n-1}}}$
Hence the ${{n}^{th}}$ term of the sequence is $\dfrac{{{2}^{n}}-1}{{{2}^{n-1}}}$.

Note: Note that while writing after getting the general term for the sequence always verify the answers by checking if the given terms match with their respected positions. For example if we substitute n = 1 we get 1, for n = 2 we get $\dfrac{3}{2}$ and for n = 3 we get $\dfrac{15}{8}$ .