A.MATHS MI 問題

When n = 1, LHS = (1+1)^2 = 4
RHS = (1)(3)(8)/6 = 24/6 = 4
The statement is true for n = 1.
Assume the statement is true for n = k,
i.e. (k+1)^2 + (k+2)^2 + (k+3)^2 +...+ (2k)^2 = [k(2k+1)(7k+1)]/6
When n = k+1,
LHS = (k+2)^2 + (k+3)^2 + (k+4)^2 +...+ (2k)^2 + (2k+1)^2 + (2k+2)^2
= (k+1)^2 + (k+2)^2 + (k+3)^2 +...+ (2k)^2 + (2k+1)^2 + 4(k+1)^2 - (k+1)^2
= [k(2k+1)(7k+1)]/6 + (2k+1)^2 + 3(k+1)^2
= (14k^3 + 9k^2 + k)/6 + (4k^2 + 4k + 1) + (3k^2 + 6k + 3)
= (14k^3 + 51k^2 + 61k + 24)/6
= (k+1)(2k+3)(7k+8)/6 [***By long division]
= (k+1)[2(k+1)+1][7(k+1)+1] /6
The statement is true for n = k+1.
By M.I., the statement is true for all natural numbers n.

trivial for n=1.
for n=k, assume it is true for some integers k>=1
when n=k+1,
(k+2)^2+(k+3)^2+...+(2k+2)^2=
((k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2 )+ [(2k+1)^2+(2k+2)^2-(k+1)^2]=
k(2k+1)(7k+1)/6+ [4k^2+4k^2+4k+8k+1+4-(k^2+2k+1)]=
k(2k+1)(7k+1)/6+ [7k^2+10k+4]=
k(2k+1)(7k+1)/6+ [42k^2+60k+24]/6=
[k(14k^2+9k+1)+(42k^2+60k+24) ]/6=計加數= [14k^3+51k^2+61k+21 ]/6=
( k+1)(2k+3)(7k+8)/6
which is also true for n=k+1.
By MI, it is always true that the preposition(qs) is true.