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.
Answer
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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.
Last updated date: 26th Sep 2023
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