Question

By the principle of mathematical induction, prove that for \[n\ge 1\], \[{{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.......+{{\left( 2n-1 \right)}^{2}}=\dfrac{n\left( 2n-1 \right)\left( 2n+1 \right)}{3}\].

Answer 1

Hint: As the given expression as P (n). Prove that the expression is true for n =1. Now, assume that P (n) is true for n = m and in the next prove that P (n) is true for, n = m + 1. Once P (n) is proved true for, n = m + 1 then we can say that the expression: - \[{{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.......+{{\left( 2n-1 \right)}^{2}}=\dfrac{n\left( 2n-1 \right)\left( 2n+1 \right)}{3}\] is true for all \[n\ge 1\].

Complete step by step answer:
Hence, we have to prove that: - \[{{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.......+{{\left( 2n-1 \right)}^{2}}=\dfrac{n\left( 2n-1 \right)\left( 2n+1 \right)}{3}\] is true for all \[n\ge 1\], using the principle of mathematical induction.
Let us assume the given expression as P (n).
\[\Rightarrow P\left( n \right)={{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.......+{{\left( 2n-1 \right)}^{2}}=\dfrac{n\left( 2n-1 \right)\left( 2n+1 \right)}{3}\]
Let us check whether P (n) is true for n = 1 or not. So, for n = 1, we have,
L.H.S. = \[{{1}^{2}}=1\]
R.H.S. = \[\dfrac{1\left( 2\times 1-1 \right)\left( 2\times 1+1 \right)}{3}=\dfrac{3}{3}=1\]
Hence, P (n) is true for n = 1.
Now, assuming that P (n) is true for n = m, we have,
\[\Rightarrow {{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.......+{{\left( 2m-1 \right)}^{2}}=\dfrac{m\left( 2m-1 \right)\left( 2m+1 \right)}{3}\] - (1)
Now let us check for, n = m + 1.
L.H.S = \[{{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.......+{{\left( 2m-1 \right)}^{2}}+{{\left( 2m+1 \right)}^{2}}\]
Using relation (i), we have,
L.H.S. = \[\dfrac{m\left( 2m-1 \right)\left( 2m+1 \right)}{3}+{{\left( 2m+1 \right)}^{2}}\]
\[\begin{align}
  & \Rightarrow L.H.S.=\dfrac{\left( 2m+1 \right)}{3}\left[ m\left( 2m-1 \right)+\left( 2m+1 \right)\times 3 \right] \\
 & \Rightarrow L.H.S.=\dfrac{\left( 2m+1 \right)}{3}\left[ 2{{m}^{2}}-m+6m+3 \right] \\
 & \Rightarrow L.H.S.=\dfrac{\left( 2m+1 \right)}{3}\left[ 2{{m}^{2}}+5m+3 \right] \\
 & \Rightarrow L.H.S.=\dfrac{\left( 2m+1 \right)\left( m+1 \right)\left( 2m+3 \right)}{3} \\
 & \Rightarrow R.H.S.=\dfrac{\left( m+1 \right)\left( 2m+2-1 \right)\left( 2m+2+1 \right)}{3} \\
 & \Rightarrow R.H.S.=\dfrac{\left( m+1 \right)\left( 2m+1 \right)\left( 2m+3 \right)}{3} \\
\end{align}\]
Clearly, we can see that for n = m + 1, L.H.S. = R.H.S, so P (n) is true for n = m + 1.
Hence, P (n) will be true for all \[n\ge 1\]. Hence proved.

Note: One may note that we have to follow the basic approach to solve a question by mathematical induction. First we need to prove the result for n = 1, then assume that it is true for n = m and then again prove that P (n) is true for n = m + 1, using the assumed relation. If any of the steps is missed then the proof is considered as incomplete proof.