Question

Let\[f(n)\]equals to the sum of the cubes of three consecutive natural numbers. \[f(n)\]leaves the remainder zero when divided by
A. 11
B. 9
C. 99
D. None of these

Answer 1

Hint: We are given \[f(n)\]equals to the sum of the cubes of three consecutive natural numbers, so we first form an equation. Let us use the concept of mathematical induction and find the value of \[f(1)\], \[f(k)\]and \[f(k + 1)\]. Hence from the calculation of three, we can get the idea about a particular number divisible from the number and leaves remainder zero. Hence, a solution can be obtained from there.

Complete step by step answer:

As the given is \[f(n)\] equals to the sum of the cubes of three consecutive natural numbers.
Hence, \[f\left( n \right) = {\left( {n - 1} \right)^3} + {n^3} + {\left( {n + 1} \right)^3}\]
Hence, expand the simplify the function using the formula
\[{\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\] and \[{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\]
Hence, on expanding, we get,
\[{\left( {n - 1} \right)^3} = {n^3} - 3{n^2} + 3n - 1\]
\[{\left( {n + 1} \right)^3} = {n^3} + 3{n^2} + 3n + 1\]
Hence, put the above value in\[f\left( n \right)\].
\[f\left( n \right) = {n^3} - 3{n^2} + 3n - 1 + {n^3} + {n^3} + 3{n^2} + 3n + 1\]
Hence, on simplification, we get,
\[ \Rightarrow \] \[f\left( n \right) = 3{n^3} + 6n\]
For \[n = 1\]
Hence, on calculating, we get,
\[ \Rightarrow \] \[f\left( 1 \right) = 3 + 6 = 9\]
We can say that \[f(n)\] is divisible by 9.
For \[n = k\],
We have,\[f\left( k \right) = 3{k^3} + 6k\]
We assume that \[f(k)\] is divisible by 9.
For \[n = k + 1\]
Now, calculating \[f\left( {k + 1} \right)\]
\[f\left( {k + 1} \right) = 3{\left( {k + 1} \right)^3} + 6\left( {k + 1} \right)\]
Expand using the formula as \[{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\].
Hence, \[f\left( {k + 1} \right) = 3\left( {{k^3} + 3{k^2} + 3k + 1} \right) + 6\left( {k + 1} \right)\]
On expanding, we get,
\[ \Rightarrow \]\[f\left( {k + 1} \right) = 3{k^3} + 6k + 9\left( {{k^2} + k + 1} \right)\]
Now, we can mould it to
\[ \Rightarrow \]\[f\left( {k + 1} \right) = f\left( k \right) + 9\left( {{k^2} + k + 1} \right)\]
Now as we had \[f(k)\] is divisible by 9 and as we can see that the second term is also a multiple of 9,
So, from here we can say that \[f\left( {k + 1} \right)\]is divisible by \[9\].
Hence, from the principle of mathematical induction, \[f\left( n \right)\]is divisible by 9 for all \[n \in 9\]
Hence, option (B) is correct answer.

Note: Mathematical Induction is a technique of proving a statement, theorem or formula which is thought to be true, for each and every natural number n. By generalizing this in form of a principle which we would use to prove any mathematical statement is 'Principle of Mathematical Induction'.
Definition. Mathematical Induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number. The technique involves two steps to prove a statement, as stated below − Step 1(Base step) − It proves that a statement is true for the initial value.
Use the formula \[{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\]and \[{\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\]properly and thus following value can be determined.