Induction method is used to prove a statement. Most commonly, it is used to prove a statement, involving, say #n# where #n# represents the set of all natural numbers.
Induction method involves two steps, One, that the statement is true for #n=1# and say #n=2#. Two, we assume that it is true for #n=k# and prove that if it is true for #n=k#, then it is also true for #n=k+1#.
First Step #-# Now for #1^2+3^2+5^2+...+(2n-1)^2=n/3(4n^2-1)#, we know for #n=1#, we have #1^2=1/3(4*1^2-1)=1/3xx3=1# and for #n=2#, we have #1^2+3^2=2/3(4*2^2-1)=2/3xx15=10#.
Hence, given statement is true for #n=1# and #n=2#.
Second Step #-# Assume it is true for #n=k#, hence
#1^2+3^2+5^2+...+(2k-1)^2=k/3(4k^2-1)#
Now let us test it for #n=k+1# i.e.
#1^2+3^2+5^2+...+(2k-1)^2+(2k+1)^2#
= #k/3(4k^2-1)+4k^2+4k+1#
= #k/3(4k^2+8k+4-1)-k/3(8k+4)+4k^2+4k+1#
= #k/3(4k^2+8k+4-1)-(8k^2)/3-(4k)/3+4k^2+4k+1#
= #(k+1)/3(4k^2+8k+4-1)-1/3(4k^2+8k+4-1)-(8k^2)/3-(4k)/3+4k^2+4k+1#
= #(k+1)/3(4k^2+8k+4-1)-color(red)((4k^2)/3)-color(blue)((8k)/3)-1-color(red)((8k^2)/3)-color(blue)((4k)/3)+color(red)(4k^2)+color(blue)(4k)+1#
= #(k+1)/3(4(k+1)^2-1)#
Hence we see that the statement is true for #n=k+1# if it is true for #n=k#.
Hence #1^2+3^2+5^2+...+(2n-1)^2=n/3(4n^2-1)# is true for all values of #ninNN#
