Answer with Step-by-step explanation:
Since we have given that
1+4+9+........................+n² = [tex]\dfrac{n(n+1)(2n+1)}{6}[/tex]
We will show it using induction on n:
Let n = 1
L.H.S. :1 = R.H.S. : [tex]\dfrac{1\times 2\times 3}{6}=\dfrac{6}{6}=1[/tex]
So, P(n) is true for n = 1
Now, we suppose that P(n) is true for n = k.
[tex]1+4+9+...................+k^2=\dfrac{k(k+1)(2k+1)}{6}[/tex]
Now, we will show that P(n) is true for n = k+1.
So, it L.H.S. becomes,
[tex]1+4+9+......................+(k+1)^2[/tex]
and R.H.S. becomes,
[tex]\dfrac{(k+1)(k+2)(2k+3)}{6}[/tex]
Consider, L.H.S.,
[tex]1+4+9+..+k^2+(k+1)^2\\\\=\dfrac{k(k+1)(2k+1)}{6}+(k+1)^2\\\\=k+1[\dfrac{k(2k+1)}{6}+(k+1)]\\\\=(k+1)[\dfrac{2k^2+k+6k+6}{6}]\\\\=(k+1)\dfrac{2k^2+7k+6}{6}]\\\\=(k+1)\dfrac{2k^2+4k+3k+6}{6}]\\\\=(k+1)[\dfrac{2k(k+2)+3(k+2)}{6}]\\\\=\dfrac{(k+1)(2k+3)(k+2)}{6}[/tex]
So, L.H.S. = R.H.S.
Hence, P(n) is true for all integers n.
