How to calculate 1 04 to the power of 45

I'm guessing you're asking how you can get an approximate result quickly, otherwise you just need to press the calculator.

By the power series: (1+x) n = 1 + nx + n(n-1) x 2 (2!).) n(n-1)(n-2)*x^3/(3!)

When x is very small, take the first few terms to get a better approximate result. For your question, ( = ( = (1+ *10 (-90)], now, find (1+x) 45 for x=.

If we take the first two terms of the power series, we get: (1+x) 45 is approximately equal to 1+45* = 1 + =

If we take the first 3 terms of the power series, we get: (1+x) 45 is approximately equal to + 45*44* = + =

If we take the first 4 terms of the power series, we get: (1+x) 45 is approximately equal to + 45*44*43*( = + =

If we take the first 5 terms of the power series, we get: (1+x) 45 is approximately equal to + 45*44*43*42*( 24 = + =

And so on until you want the accuracy. You may find that the above calculations converge slowly, because x is not small enough, and n=45 is still relatively large.

If there is such a question: for example, every year grows and grows for 45 years, how much will the result be? Then solve (1+, which is very small, and only need to take the first 2 terms of the power series, (1+ is about 1+45* = .

5 to the power.

In fact, it is 5 * 5 of 2 5 to the 5th power of the boy.

That is, 5 * (25 of 5 times the square root number).

You can use a calculator to get the grandchildren. Approximately equal to.

Summary. Hello, Mr. Jiang Qiusheng is at your service, I have seen your question, please give me 1-5 minutes, I will sort out the answer and send it to you, thank you for your patience.

Hello, Mr. Jiang Qiusheng is at your service, I have read and checked your question, please give me 1-5 minutes, I will sort out the answer and send it to you after Hong's La, thank you for your patience.

of the 40th power, this calculation is the multiplication of 40, and the result of the calculation is: Mu Oak needs to use the calculation of Tannai socks to calculate the excitation.

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The agitation of the 43rd power high lead pickpocketing Qi Chang.

The calculation is as follows: The 35th power of Yu Seri He is:

This is a multi-power calculation problem for decimals, and it is used by the Yunqing faction to calculate the first Xun machine.

Summary. Let the common ratio of the proportional series {an} be q, and the sum of the first n terms is sn

sn=a1+a2+a3+……a(n-1)+an

a1+a1*q+a1*q^2+……a1*q^(n-2)+a1*q^(n-1)

Multiply both sides of the equation by the common ratio q

q*sn=a1*q+a1*q^2+a1*q^3+……a1*q^(n-1)+a1*q^n

Subtract the two formulas. sn-q*sn

a1+(a1*q-a1*q)+(a1*q^2-a1*q^2)+…a1*q^(n-1)-a1*q^(n-1)]-a1*q^n

a1-a1*q^n

i.e. (1-q)*sn=a1*(1-q n).

sn=a1*(1-q n) (1-q).

Specific to the title of the landlord.

f=100*[1+(1+

It can be seen that in the middle brackets is the sum of the first 4 terms of the proportional number series with the first term 1 and the common ratio of 1+.

Apply the above formula, a1=1, q=1+, n=4, to obtain.

f=100*

The formula for calculating 1+ to the 3rd power.

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Let the common ratio of the proportional series {an} be q, and the sum of the first n terms is snsn=a1+a2+a3+......a(n-1)+an=a1+a1*q+a1*q^2+……Both sides of the equation a1*q (n-2)+a1*q (n-1) refer to the common ratio qq*sn=a1*q+a1*q 2+a1*q 3+......a1*q (n-1)+a1*q n subtract sn-q*sn=a1+(a1*q-a1*q)+(a1*q 2-a1*q 2)+....a1*q (n-1)-a1*q (n-1)]-a1*q n=a1-a1*q n, that is, (1-q)*sn=a1*(1-q n) to get sn=a1*(1-q n) (1-q) specific to the landlord's topic f=100*[1+(1+It can be seen that in parentheses is the first item is 1, and the first 4 items of the proportional number series with a reputation ratio of 1+, the sum of the first 4 items is applied to the formula of the upper Qin section surface, a1=1, q=1+, n=4, f=100*=100*[( 1+