標題:
sum of 11th term
發問:
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(-5)^2+(-4)^2+(-3)^2+............+11th term pls. 更新: 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6 what mean??
最佳解答:
(-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.
其他解答:
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
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