sum of 11th term

(-5)^2+(-4)^2+(-3)^2+............+11th term
= 25+16+9+4+1+0+1+4+9+16+25
= 2*(25+16+9+4+1) = 110

you may also use 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6,
but I don't think it would be useful becuase n is very small, it can be easily calculated by addition.

http://j.imagehost.org/0277/ScreenHunter_01_Jun_16_12_42.gif

1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6 是一條公式
(-5)^2+(-4)^2+(-3)^2+............+11th term
(1^2+2^2+3^2+4^2+5^2)x2
=(5x6x11/6)x2
=110

(-5)^2+(-4)^2+(-3)^2+(-2)^2+(-1)^2+...+11th term
=(-5)^2+(-4)^2+(-3)^2+(-2)^2+(-1)^2+(0)^2+(1)^2+(2)^2+(3)^2+(4)^2+(5)^2
=25+16+9+4+1+0+1+4+9+16+25
=110

Because
(-5)^2 = 5^2
(-4)^2 = 4^2
(-3)^2 = 3^2
... and so on.

Also,by Pyth.th, 4^2 + 3^2 = 5^2

Therefore,
(-5)^2+(-4)^2+(-3)^2+............+11th term
=(5^2 + 5^2 + 2^2 + 1^2) x2
=(25+25+2+1)x2
=53x2
=106

2007-06-05 12:05:43 補充：
Sorry, 更正=(25 25 4 1)x2=55x2=110