Prove by mathematical induction that (n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2=n(2n+1)(7n+1)/6, is true for all natural numbers n.
(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2=n(2n+1)(7n+1)/6,
n=1
2^2 = 1.(3)(8)/6
4 = 4
Which is true.
Therefore statement is true for n = 1
Assume statement is true for n = k.
n = k
(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2=k(2k+1)(7k+1)/6, ------------------------- (1)
n = (k+1)
(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2+(2k+2)^2=(k+1)(2k+3)(7k+8)/6,
Lhs = (k+1)^2 + (k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2 + {(2k+1)^2+(2k+2)^2 – (k+1)^2}
Lhs = k(2k+1)(7k+1)/6 + {(2k+1)^2+(2k+2)^2 – (k+1)^2} (using (1) )
Lhs = k(2k+1)(7k+1)/6 + {4k^2 + 4k + 1 + 4k^2 + 8k + 4 – k^2 – 2k - 1}
Lhs = k(2k+1)(7k+1)/6 + {7k^2 + 10k + 4}
Lhs = (14k^3 + 9k^2 + k)/6 + (42k^2 + 60k + 24)/6
Lhs = (14k^3 + 51k^2 + 61k + 24)/6
Lhs = (k+1)(14k^2 + 37k + 24)/6
Lhs = (k+1)(2k+3)(7k+8)/6 = rhs
Lhs = rhs
Hence by the process of Mathematical Induction, the statement is true for all n = 1,2,3,...