Question

By method of induction, prove that 1.3 + 2.5 + 3.7+…..+n(2n+1)=$\dfrac{n}{6}(n + 1)(4n + 5)$ for all $n \in N$

Answer 1

At first we need to prove that the result is true for P(1) and assuming the result is true for n=k we need to prove the result is true for n = k+1.If it is true then the result is true for all

Complete step-by-step answer:
To solve a problem using mathematical induction we need follow a few steps
First let find P(n)
Here P(n) = 1.3 + 2.5 + 3.7+…..+n(2n+1)=$\dfrac{n}{6}(n + 1)(4n + 5)$
Now let's find P(1)
In the left hand side
P(1) = 1.3 = 3
In the right hand side
P(1) = $\dfrac{1}{6}(1 + 1)(4(1) + 5) = \dfrac{1}{6}(2)(9) = \dfrac{{18}}{6} = 3$
Hence the result is true for P(1)
So let's assume that the result is true for all n = k
$\Rightarrow 1.3{\text{ }} + {\text{ }}2.5{\text{ }} + {\text{ }}3.7 + \ldots .. + k\left( {2k + 1} \right) = \dfrac{k}{6}(k + 1)(4k + 5)$………(1)
Now we need to prove that the result is true for n = k+1
That is we need to prove $1.3{\text{ }} + {\text{ }}2.5{\text{ }} + {\text{ }}3.7 + \ldots .. + (k + 1)\left( {2(k + 1) + 1} \right) = \dfrac{{k + 1}}{6}((k + 1) + 1)(4(k + 1) + 5)$
For that lets take the left hand side
$\Rightarrow 1.3{\text{ }} + {\text{ }}2.5{\text{ }} + {\text{ }}3.7 + \ldots .. + (k + 1)\left( {2(k + 1) + 1} \right)$
$\Rightarrow 1.3{\text{ }} + {\text{ }}2.5{\text{ }} + {\text{ }}3.7 + \ldots .. + k\left( {2k + 1} \right) + (k + 1)(2(k + 1) + 1)$
From (1) we get
$= \dfrac{k}{6}(k + 1)(4k + 5) + (k + 1)(2(k + 1) + 1) \\\ = \dfrac{k}{6}(k + 1)(4k + 5) + (k + 1)(2k + 2 + 1) \\\ = \dfrac{k}{6}(k + 1)(4k + 5) + (k + 1)(2k + 3) \\\ =(k + 1)\left[ {\dfrac{k}{6}(4k + 5) + (2k + 3)} \right] \\\ =(k + 1)\left[ {\dfrac{{k(4k + 5) + 6(2k + 3)}}{6}} \right] \\\ = (k + 1)\left[ {\dfrac{{4{k^2} + 5k + 12k + 18}}{6}} \right] \\\ = (k + 1)\left[ {\dfrac{{4{k^2} + 17k + 18}}{6}} \right] \\\$
By using splitting the middle term method
$=(k + 1)\left[ {\dfrac{{4{k^2} + 17k + 18}}{6}} \right] \\\ = (k + 1)\left[ {\dfrac{{4{k^2} + 8k + 9k + 18}}{6}} \right] \\\ =(k + 1)\left[ {\dfrac{{4k(k + 2) + 9(k + 2)}}{6}} \right] \\\ =(k + 1)\left[ {\dfrac{{(k + 2)(4k + 9)}}{6}} \right] \\\ = \dfrac{{k + 1}}{6}(k + 1 + 1)(4k + 4 + 5) \\\ =\dfrac{{k + 1}}{6}((k + 1) + 1)(4(k + 1) + 5) \\\$
Hence we have proved that the result is true for n = k+1
Therefore the result is true for all $n \in N$.

Note: Mathematical induction is a mathematical proof technique. It is essentially used to prove that a statement P(n) holds for every natural number n = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases P(0), P(1), P(2), P(3),….
we can use mathematical induction to prove that a propositional function P(n) is true for all integers n≥1