Question

Here, \[\forall n \in N;3{n^5} + 5{n^3} + 7n\] is divisible by
A. \[3\]
B. \[5\]
C. \[10\]
D. \[15\]

Answer 1

Hint: First, we need to find the given sum of algebraic expression is divisible by the above mentioned options which one is true. To find the possible division by apply mathematical induction method, let assume, \[P(n)\] \[n = 1,2...n + 1\] to prove the statement is true and compare each other and find the divisible value and recall the formula,
  \[
  {(a + b)^5} = {a^5} + 5{a^4}b + 10{a^3}{b^2} + 10{a^2}{b^3} + 5a{b^4} + {b^5} \\
  {(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} \;
 \]

Complete step by step solution:
Given algebraic expression is \[\forall n \in N;3{n^5} + 5{n^3} + 7n\] ,
By using mathematical induction,
Let us assume, \[P(n) = 3{n^5} + 5{n^3} + 7n\] , we get
Put, \[n = 1\] into the given expression
 \[P(1) = 3{(1)^5} + 5{(1)^3} + 7(1)\]
To simplify the addition, we get
 \[p(1) = 3 + 5 + 7 = 15\] is divisible by \[3,5,15\] .
If Put, \[n = 2\] into the given expression
 \[P(2) = 3{(2)^5} + 5{(2)^3} + 7(2) = 32 + 40 + 14 = 150\]
Which is divisible by \[3,5,10,15\] .
Therefore, \[P(1)\] and \[P(2)\] both are divisible by \[3,5,15\]
Every number divisible by \[15\] is divisible by \[3\] and \[5\] .
Let us assume, \[P(n) = 3{n^5} + 5{n^3} + 7n\] is divisible by \[15\] .
Then, \[3{n^5} + 5{n^3} + 7n = 15\lambda \] .
Finally, we have to prove the statement is true for \[n + 1\] , the expression is divisible by \[3,5,15\] .
Let us consider, \[P(n + 1) = 3{(n + 1)^5} + 5{(n + 1)^3} + 7(n + 1)\]
We know the formula, we have
 \[
  {(a + b)^5} = {a^5} + 5{a^4}b + 10{a^3}{b^2} + 10{a^2}{b^3} + 5a{b^4} + {b^5} \\
  {(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} \\
\]
By substitute \[{(n + 1)^5}\] and \[{(n + 1)^3}\] in the above formula, we get
 \[P(n + 1) = 3({n^5} + 5{n^4} + 10{n^3} + 10{n^2} + 5n + 1) + 5({n^3} + 3{n^2} + 3n + 1) + 7(n + 1)\]
By simplify the arithmetic operation,
 \[P(n + 1) = 3{n^5} + 15{n^4} + 30{n^3} + 30{n^2} + 15n + 3 + 5{n^3} + 15{n^2} + 15n + 5 + 7n + 7\]
Now, we get
 \[
  P(n + 1) = (3{n^5} + 5{n^3} + 7n) + 15{n^4} + 30{n^3} + 45{n^2} + 30n + 15 \\
  p(n + 1) = 15\lambda + 15({n^4} + 2{n^3} + 3{n^2} + 2n + 1) \\
 \]
Take \[15\] commonly out from the bracket,
 \[p(n + 1) = 15(\lambda + {n^4} + 2{n^3} + 3{n^2} + 2n + 1)\]
Which is divisible by 15.
Therefore, \[p(n + 1)\] is divisible by \[15\]
 By the mathematical induction, given expression is divisible by \[15\] , for \[\forall n \in N\]
Hence, the final answer is the given expression is divisible by \[3,5,15\] .
So, the correct answer is “Option D”.

Note: Here, we use mathematical induction to solve the given expression. There are three principles in the induction method is mentioned below,
Principle of Mathematical Induction: If \[P(n)\] is a given expression, \[\;n\] is a natural number
I.Verify if the statement is true for \[n = 1\] , \[P(1)\] is true. Then,
II.Assume that the statement is true for \[n = k\] , \[P(k)\] is true
III.If the statement is true for \[n = k + 1\] then the statement is true for all values of \[n\] .