Question

A pattern is given below
$1=1^2=1$
$4=2^2=1+3$
$9=3^2=1+3+5$
$16=4^2=1+3+5+7$etc.
Based on above pattern, we have
$P(n):1+3+5+7+\cdots +(2n-1)=n^2$
Based on above information, which among the following is true?

• A. $P(1)$ is true
• B. If $P(k)$ is true, then $P(k+1)$ is also true, where $k$ is some positive integer
• C. $P(k+1)=k^2+2k+2$, where $k$ is some positive integer
• D. Both (a) and (b) are true

Answer 1

Answer: (D)

From the given pattern,
$1=1^2=1$
$4=2^2=1+3$
$9=3^2=1+3+5$
$16=4^2=1+3+57$, etc.
It is worth to be noted that the sum of the first two odd natural numbers is the square of second natural number, sum of the first three odd natural numbers is the square of third natural number and so on. Thus, from this pattern, it appears that
$1+3+5+7+\dots .+(2n-1)=n^2$
i.e., the sum of the first $n$ odd natural numbers is the square of $n$.
Let us write
$P(n):1+3+5+7+\dots ..+(2n-1)-n^2.$
To prove that $P(n)$ is true for all $n$.
The first step in a proof that uses mathematical induction is to prove that $P(1)$ is true. This step is called the basic step.
Obviously
$1=1^2$, i.e., $P(1)$ is true.
The next step is called the inductive step. Here, we suppose that $P(k)$ is true for some positive integer $k$, i.e.,
$1+3+5+7+\dots +(2k-1)=k^2....$(i)
Now, we shall prove $P(k+1)$ is true.
Consider,
$1+3+5+7+\dots \dots +(2k-1)+\{2(k+1)-1\}.....$(ii)
$=k^2+(2k+1)=(k+1{)}^2$[ using Eq.(i)]
Therefore, $P(k+1)$ is true and the inductive proof is now completed. Hence, $P(n)$ is true for all natural numbers $n$.