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Question: Find the sequence if it is given that the \({{n}^{th}}\) term of the series is given by \({{T}_{n}}=...

Find the sequence if it is given that the nth{{n}^{th}} term of the series is given by Tn=2n−1{{T}_{n}}=2n-1 ?

Explanation

Solution

In this question we are given the nth{{n}^{th}} term of the series and we have to find the series using that. So first of all we will find the first term of the series by putting the n = 1 in the equation Tn=2n−1{{T}_{n}}=2n-1 and similarly we will then find the rest of the values for n = 2, 3…. and so on. After finding some initial values we will combine all the values and represent them as a series as asked in the question.

Complete step by step answer:
We are given in the question that nth{{n}^{th}} term of a series is given by Tn=2n−1{{T}_{n}}=2n-1, and we have to find the sequence,
So first of all we will find the first term of the sequence by putting the value of n = 1 in the given nth{{n}^{th}} term,
So we will do this as,
Tn=2n−1{{T}_{n}}=2n-1
T1=(2×1)−1 T1=2−1 T1=1 \begin{aligned} & {{T}_{1}}=\left( 2\times 1 \right)-1 \\\ & {{T}_{1}}=2-1 \\\ & {{T}_{1}}=1 \\\ \end{aligned}
Similarly we will find the second term by putting n = 2,
Tn=2n−1 T2=(2×2)−1 T2=4−1 T2=3 \begin{aligned} & {{T}_{n}}=2n-1 \\\ & {{T}_{2}}=\left( 2\times 2 \right)-1 \\\ & {{T}_{2}}=4-1 \\\ & {{T}_{2}}=3 \\\ \end{aligned}
Now third term, we will find in a similar manner,

& {{T}_{n}}=2n-1 \\\ & {{T}_{3}}=\left( 2\times 3 \right)-1 \\\ & {{T}_{3}}=6-1 \\\ & {{T}_{3}}=5 \\\ \end{aligned}$$ If we observe we can see that the formed series is in the arithmetic progression and it common difference is, $\begin{aligned} & ={{T}_{n+1}}-{{T}_{n}} \\\ & =2\left( n+1 \right)-1-\left( 2n-1 \right) \\\ & =2n+2-1-2n+1 \\\ & =2 \\\ \end{aligned}$ So by adding 2 to the previous term we can find the next term of this series. Hence we get the sequence as, 1, 3, 5, ……… **Note:** Whenever some series is given in the questions then try to carefully observe the question first and then try to solve it as in this question if we would have observed the sequence carefully before starting solving it then we may have found that it is a Arithmetic progression series with common difference 2 and then the terms of the series could have been easily found out by just adding 2 to the first term.