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Question: By the principle of mathematical induction, prove that for \[n\ge 1\], \[{{1}^{2}}+{{3}^{2}}+{{5}^{2...

By the principle of mathematical induction, prove that for n≥1n\ge 1, 12+32+52+.......+(2n−1)2=n(2n−1)(2n+1)3{{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.......+{{\left( 2n-1 \right)}^{2}}=\dfrac{n\left( 2n-1 \right)\left( 2n+1 \right)}{3}.

Explanation

Solution

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: - 12+32+52+.......+(2n−1)2=n(2n−1)(2n+1)3{{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≥1n\ge 1.

Complete step by step answer:
Hence, we have to prove that: - 12+32+52+.......+(2n−1)2=n(2n−1)(2n+1)3{{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≥1n\ge 1, using the principle of mathematical induction.
Let us assume the given expression as P (n).
⇒P(n)=12+32+52+.......+(2n−1)2=n(2n−1)(2n+1)3\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. = 12=1{{1}^{2}}=1
R.H.S. = 1(2×1−1)(2×1+1)3=33=1\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,
⇒12+32+52+.......+(2m−1)2=m(2m−1)(2m+1)3\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 = 12+32+52+.......+(2m−1)2+(2m+1)2{{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.......+{{\left( 2m-1 \right)}^{2}}+{{\left( 2m+1 \right)}^{2}}
Using relation (i), we have,
L.H.S. = m(2m−1)(2m+1)3+(2m+1)2\dfrac{m\left( 2m-1 \right)\left( 2m+1 \right)}{3}+{{\left( 2m+1 \right)}^{2}}

& \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{aligned}$$ Clearly, we can see that for n = m + 1, L.H.S. = R.H.S, so P (n) is true for n = m + 1. Hence, P (n) will be true for all $$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.