# Show that every positive even integer is of the form 2n and every positive odd integer is of the form 2n+1.

Last updated date: 27th Mar 2023

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Answer

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Hint: We have to show that every positive even integer is of the form 2n and every positive odd integer is of the form 2n+1. In order to prove this, form a series of all the positive even natural numbers and in the same manner form a series of all the positive natural odd numbers. Use this series to understand that whether it forms an A.P, G.P or H.P. Then evaluate the general ${n^{th}}$ of this series formulated to get the proof.

Complete step-by-step answer:

$\left( i \right)$Let us consider positive even terms, as we know that even terms are always divisible by 2.

Therefore the set of positive even terms are $\left( {2,4,6,8,........................} \right)$ up to n terms.

Now as we see this forms an A.P with first term (${a_1}$) = 2, common difference (d) = (4-2) = (6-4) = 2.

Now we have to find out the last term (${a_n}$) of this A.P

So use the formula of ${n^{th}}$ term of an A.P which is ${a_n} = {a_1} + \left( {n - 1} \right)d$

${a_n} = 2 + \left( {n - 1} \right)2 = 2 + 2n - 2 = 2n$

Therefore every positive even integer is of the form of 2n.

$\left( {ii} \right)$Let us consider positive odd terms, as we know that odd terms are not divisible by 2.

Therefore the set of positive odd terms are $\left( {1,3,5,7........................} \right)$ up to n terms.

Now as we see this forms an A.P with first term (${a_1}$) = 1, common difference (d) = (3-1) = (5-3) = 2.

Now we have to find out the last term (${a_n}$) of this A.P

So use the formula of ${n^{th}}$ term of an A.P which is ${a_n} = {a_1} + \left( {n - 1} \right)d$

${a_n} = 1 + \left( {n - 1} \right)2 = 1 + 2n - 2 = 2n - 1$

Therefore every positive odd integer is of the form of (2n-1).

Note: Whenever we face such types of problems the key concept involved in formulation of the series to evaluate the ${n^{th}}$ term. In many questions we may need these results directly, let us suppose we have to find two consecutive even or odd natural numbers then the direct application of this proved result about the general even term as 2n and odd term as 2n+1, will help you reach the right answer. So always keep this result in mind.

Complete step-by-step answer:

$\left( i \right)$Let us consider positive even terms, as we know that even terms are always divisible by 2.

Therefore the set of positive even terms are $\left( {2,4,6,8,........................} \right)$ up to n terms.

Now as we see this forms an A.P with first term (${a_1}$) = 2, common difference (d) = (4-2) = (6-4) = 2.

Now we have to find out the last term (${a_n}$) of this A.P

So use the formula of ${n^{th}}$ term of an A.P which is ${a_n} = {a_1} + \left( {n - 1} \right)d$

${a_n} = 2 + \left( {n - 1} \right)2 = 2 + 2n - 2 = 2n$

Therefore every positive even integer is of the form of 2n.

$\left( {ii} \right)$Let us consider positive odd terms, as we know that odd terms are not divisible by 2.

Therefore the set of positive odd terms are $\left( {1,3,5,7........................} \right)$ up to n terms.

Now as we see this forms an A.P with first term (${a_1}$) = 1, common difference (d) = (3-1) = (5-3) = 2.

Now we have to find out the last term (${a_n}$) of this A.P

So use the formula of ${n^{th}}$ term of an A.P which is ${a_n} = {a_1} + \left( {n - 1} \right)d$

${a_n} = 1 + \left( {n - 1} \right)2 = 1 + 2n - 2 = 2n - 1$

Therefore every positive odd integer is of the form of (2n-1).

Note: Whenever we face such types of problems the key concept involved in formulation of the series to evaluate the ${n^{th}}$ term. In many questions we may need these results directly, let us suppose we have to find two consecutive even or odd natural numbers then the direct application of this proved result about the general even term as 2n and odd term as 2n+1, will help you reach the right answer. So always keep this result in mind.

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