Find the sum of the first 60 items of the arithmetic sequence with the first 13 and the tolerance 5 Solutions without unknowns

Find the sum of the first 60 items of the arithmetic sequence with the first 13 and the tolerance 5 Solutions without unknowns

The last term is 13 + 5 * (60-1) = 308
The sum is (13 + 308) * 60 / 2 = 9630

Find the sum of the first 1999 terms of the arithmetic sequence with the first term of 5 and the tolerance of 3?

an＝a1＋（n－1）d
sn＝（a1＋an）×n/2
＝a1×n/2＋（n－1）d×n/ 2
The results show that s1999 = 6000998

The first term is the sum of the first 30 terms of the arithmetic sequence with tolerance of 3.5

an=3+(n-1)*5
So A30 = 3 + 29 * 5 = 148
Sn=(3+148)*30/2=2265

If x + y = m, xy = n, then x2 + Y2=______ ，（x-y）2=______ ，x2-xy+y2=______ ．

∵ x + y = m, ∵ (x + y) 2 = m2, that is, X2 + Y2 + 2XY = m2, ∵ x2 + y2 = m2-2xy = m2-2n; (X-Y) 2 = x2 + y2-2xy = m2-2n-2n = m2-4n; x2-xy + y2 = x2 + y2-xy = m2-2n-n = m2-3n

Xy = m, and 1 / x2 + 1 / y2 = n, then (X-Y) 2=
3Y = x + 2Z, then x2 + 9y2 + 4z2-6xy-12yz + 4XZ=

(x^2+y^2)/(xy)^2=n
x^2+y^2=m^2n
(x-y)^2=m^2n-2m

Given XY > 0, compare the size of (x ^ 2 + y ^ 2) (X-Y) and (x ^ 2-y ^ 2) (x + y)
The subtraction of this formula is 2XY (Y-X), because we don't know the positive and negative size of XY. Is it necessary to discuss it by category? If so, how can we discuss it by category?

We need to discuss (x ^ 2 + y ^ 2) (X-Y) - (x ^ 2-y ^ 2) (x + y) = (x ^ 2 + y ^ 2) (X-Y) (x + y) ^ 2 = (X-Y) [(x ^ 2 + y ^ 2) - (x + y) ^ 2] = (X-Y) (x ^ 2 + y ^ 2-x ^ 2-y ^ 2-2xy) = - 2XY (X-Y) from XY > 0 when x > y, that is, X-Y > 0, the original formula = - 2XY (X-Y) 0 when x > y

Comparison size: x ^ 5 + y ^ 5 and x ^ 4 + XY ^ 4

X ^ 5 + y ^ 5 - (x ^ 4Y + XY ^ 4) = x ^ 4-x ^ 4Y + y ^ 5-xy ^ 4 = x ^ 4 (X-Y) + y ^ 4 (Y-X) = (X-Y) (x ^ 4-y ^ 4) = (X-Y) (x ^ 2 + y ^ 2) (x ^ 2-y ^ 2) = (X-Y) (x ^ 2 + y ^ 2) (x + y) (X-Y) = (X-Y) ^ 2 (x + y ^ 2) > 0, so x ^ 5 + y ^ 5 > x ^ 4Y + XY ^ 4

Try to compare the size of x ^ 3 + y ^ 3 and x ^ 2Y + XY ^ 2

Poor performance:
(x^3+y^3)-(x^2y+xy^2)
=(x^3-x^2y)+(y^3-xy^2)
=x^2(x-y)+y^2(y-x)
=(x+y)(x-y)^2
If x = y or x = - y
Then x ^ 3 + y ^ 3 = x ^ 2Y + XY ^ 2
If x > - y and X ≠ y
Then x ^ 3 + y ^ 3 > x ^ 2Y + XY ^ 2
If x

We know that | x + 2005 / 2006 | + | y + 2004 / 2005 | = 0, and compare the size of XY
It is known that | x + 2005 / 2006 | + | y + 2004 / 2005 | = 0,

According to the absolute value is non negative
x+2005/2006=0 y+2004/2005=0
x=-2005/2006 y=-2004/2005
∵2005/2006>2004/2005
∴-2005/2006

The quadratic equation x-2kx + 1 / 2k-2 = 0 with respect to X is known. The proof is that no matter what the value of K is, the equation always has two unequal real roots

In other words, the equation of one variable quadratic equation x-2kx + 1 / 2k-2 = 0 no matter what the value of K is. The equation always has two unequal real roots