1. Calculation: X & ﹥ 8226; X & ﹥ 178; - ﹥ 8226; X & ﹥ 179; - ﹥ 8226 X ^ 2009, and think about what will happen if you change 2009 to n?

1. Calculation: X & ﹥ 8226; X & ﹥ 178; - ﹥ 8226; X & ﹥ 179; - ﹥ 8226 X ^ 2009, and think about what will happen if you change 2009 to n?

Original formula = x ^ (1 + 2 + 3 +,, + 2009)
=x^(2019045)
If you change it to N, the sum of the equal differences will make the
1+2.+n=（n+1）n/2
Original formula = x × x ^ 2,,,,, x ^ n
=x^【n（n+1）/2】
Do not understand, please ask, have a good time

2 & # 178; - 1 & # 178; = 3 3 & # 178; - 2 & # 178; = 5 find the nth formula and calculate 1 + 3 + 5 + 7 +. + 2009 + 2011 with the rule

4^2-3^2=7
.
n^2-(n-1)^2=2n-1
1+3+5+7+.+2009+2011
=1^2+(2²-1²)+(3²-2²)+(4^2-3^2)+.+(1005^2-1004^2)+(1006^2-1005^2)
=1006^2
=1012036

Given 1 / √ X - √ x = - 1, find 1 / X & # 178; + X & # 178;

1/√x-√x=-1
（1/√x-√x）²=（-1）²
1/x-2+x=1
1/x+x=3
(1/x+x)²=3²
1/x²+2+x²=9
1/x²+x²=7
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