The zeroth power of 5 plus the first power of 5. The 100th power of 5 =? Explain ideas

The zeroth power of 5 plus the first power of 5. The 100th power of 5 =? Explain ideas

Let s = 5 ^ 0 + 5 ^ 1 + 5 ^ 2... + 5 ^ 99 + 5 ^ 100 1
Then 5S = 5 ^ 1 + 5 ^ 2 + 5 ^ 3... + 5 ^ 100 + 5 ^ 1012
1-2
-4S=5^0+0+0+...-5^101
So s = (5 ^ 101-1) / 4

1 / 0 of 2 + 1 / 1 of 2 + 1 / 3 of 2 + ······ + 1 / 100 of 2 =?

1+1/2+1/4+.1/2^100=(1-1/2^101)/(1-1/2)=2-1/2^100
This is an equal ratio sequence. The equal ratio is 1 / 2

The nth power of a is () where a is () and N is ()

The nth power of a is () where a is called (base) and N is called (index)

What is the 19th power of 10?

The 19th power of 10 equals 19 zeros after 1

Find the maximum value of y = 2cos square x + 5sinx-4, and find the corresponding x value at this time

Y=2(1-sinX^2)+5sinx-4 =-2(SINX-5/4)^2+2*25/16-2
When SiNx = 1, the maximum value of Y is 1

Try to find the value of y = 2cos & sup2; + 5sinx-4, and find the corresponding x value at this time
Add 80 points to the solution

There is something wrong with your subject
How to solve a function? At least it must be an equation!
And what's behind your cos? What's a cos
In addition, you are only one level, where is 80?

Finding the maximum value of the function y = 2cos ^ 2 * x + 5sinx-4

y=2-2sin²x+5sinx-4
Let t = SiNx
Then y = 2T & # 178; + 5t-4
When t = - 5 / 4 (t ∈ [- 1.1], so it is not advisable)
Then substitute t = - 1, t = 1 to get the maximum value respectively
Maximum: 3, minimum: - 7

The function f (x) = 2sinxcosx-2cos square x + 1 is known;
(2) If f (θ) = 0.6, find the value of Cos2 (π / 4-2 θ)

(1) Function f (x) = 2sinxcosx-2cos square x + 1 = sin2x-cos2x = root sign 2Sin (2x - π / 4), maximum root sign 2, when 2x - π / 4 = π / 2, x = 3 π / 8. (2) f (θ) = 0.6, sin (2 θ - π / 4) = 3, root sign 2 / 10. Cos2 (π / 4-2 θ) = Cos2 (2 θ - π / 4) = 9 / 10 or - 9 / 10

The function f (x) = 2sinxcosx-2cos square x + 1 is known
Given the function f (x) = the square of 2sinxcosx-2cos + 1 (1), find the minimum positive period of F (x)

F (x) = sin2x-co2x = asin (2x-45), a is the square root of 2, from which the minimum positive period T = 180 can be obtained

Given the function f (x) = 2cos2x + 2sinxcosx. (1) find the value of F (π 8); (2) find the minimum positive period and minimum value of F (x)

(1) F (x) = cos2x + 1 + sin2x = 2Sin (2x + π 4) + 1, (6 points)  f (π 8) = 2Sin (π 4 + π 4) + 1 = 2 + 1. (8 points) (2) from (1), it can be seen that f (x) = 2Sin (2x + π 4) + 1, the minimum positive period of function f (x) t = 2, π 2 = π. (10 points) the minimum value of function f (x) is 1 − 2. (12 points)