# Write cubes of 5 natural numbers which are of form 3n+1 (e.g.4,7,10…) and verify the following;

The cube of a natural number of the form 3n+1 is a natural number of form 3n+1.

Last updated date: 16th Mar 2023

•

Total views: 304.2k

•

Views today: 2.84k

Answer

Verified

304.2k+ views

Hint: We can find the five natural numbers of form 3n+1 by substituting 1,2,3,4,5 respectively in n and in this question 3n+1 implies that when we divide a number with 3 the remainder would be 1.

Complete step-by-step answer:

We have to write cubes of 5 natural numbers which are of form 3n+1 and we have to verify their cubes are natural numbers of form 3n+1.

The five natural numbers which are of form 3n+1 are 4,7,10,13 and 16 we obtain them by simply substituting 1,2,3,4,5 respectively in 3n+1.

The first thing we are going to do is writing cubes of 5 natural numbers (i.e. 4,7,10,13,17).

The cubes of 5 natural numbers (i.e. 4,7,10,13,16) are 64,343,1000,2197 and 4096 respectively.

The second thing we are going to do is expressing their cubes in the form 3n+1 if possible.

So, the cubes of natural numbers 4,7,10,13,17 can be expressed as:

\[64=3\times 21+1\],64 is expressed as 3n+1 where n=21.

\[343=3\times 114+1\], 343 is expressed as 3n+1 where n=114.

\[1000=3\times 333+1\], 1000 is expressed as 3n+1 where n=333.

\[2197=3\times 732+1\], 2197 is expressed as 3n+1 where n=732.

\[4096=3\times 1365+1\], 4096 is expressed as 3n+1 where n=1365.

So, here we expressed cubes of numbers 4,7,10,13,16 of form 3n+1 in the form of 3n+1.

So, the statement “The cube of natural numbers of the form 3n+1 is a natural number of the form 3n+1.” is verified.

Note: Now we will verify the above question in an alternative way, we directly cube 3n+1. We use

\[{{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\] formula and proceed.

Now cubing 3n+1, \[{{(3n+1)}^{3}}=27{{n}^{3}}+3{{n}^{2}}+3n+1\] , taking 3 common from the coefficients of \[{{n}^{3}},{{n}^{2}}\] and $n$ .

Now it becomes,

\[\Rightarrow 3(9{{n}^{3}}+{{n}^{2}}+n)+1\to (1)\]

We assume \[9{{n}^{3}}+{{n}^{2}}+n=k\]

Equation \[\text{(1)}\] becomes \[3k+1\]

Hence the statement “The cube of natural number of the form 3n+1 is a natural number of same

form.” is verified.

Complete step-by-step answer:

We have to write cubes of 5 natural numbers which are of form 3n+1 and we have to verify their cubes are natural numbers of form 3n+1.

The five natural numbers which are of form 3n+1 are 4,7,10,13 and 16 we obtain them by simply substituting 1,2,3,4,5 respectively in 3n+1.

The first thing we are going to do is writing cubes of 5 natural numbers (i.e. 4,7,10,13,17).

The cubes of 5 natural numbers (i.e. 4,7,10,13,16) are 64,343,1000,2197 and 4096 respectively.

The second thing we are going to do is expressing their cubes in the form 3n+1 if possible.

So, the cubes of natural numbers 4,7,10,13,17 can be expressed as:

\[64=3\times 21+1\],64 is expressed as 3n+1 where n=21.

\[343=3\times 114+1\], 343 is expressed as 3n+1 where n=114.

\[1000=3\times 333+1\], 1000 is expressed as 3n+1 where n=333.

\[2197=3\times 732+1\], 2197 is expressed as 3n+1 where n=732.

\[4096=3\times 1365+1\], 4096 is expressed as 3n+1 where n=1365.

So, here we expressed cubes of numbers 4,7,10,13,16 of form 3n+1 in the form of 3n+1.

So, the statement “The cube of natural numbers of the form 3n+1 is a natural number of the form 3n+1.” is verified.

Note: Now we will verify the above question in an alternative way, we directly cube 3n+1. We use

\[{{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\] formula and proceed.

Now cubing 3n+1, \[{{(3n+1)}^{3}}=27{{n}^{3}}+3{{n}^{2}}+3n+1\] , taking 3 common from the coefficients of \[{{n}^{3}},{{n}^{2}}\] and $n$ .

Now it becomes,

\[\Rightarrow 3(9{{n}^{3}}+{{n}^{2}}+n)+1\to (1)\]

We assume \[9{{n}^{3}}+{{n}^{2}}+n=k\]

Equation \[\text{(1)}\] becomes \[3k+1\]

Hence the statement “The cube of natural number of the form 3n+1 is a natural number of same

form.” is verified.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE