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

Answer
Verified

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.

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