ilorty

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Имя пользователяilorty
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 #1
avatar+1094 
+1

Didn't you just ask this like 3 minutes ago? frown

 

Also, it has been answered here. But, finding that you probably don't have the time to go over to the link, as you are too busy posting questions every 3 minutes, here is the answer guest has posted:

 

(a) 101^2 - 97^2 + 93^2 - 89^2 + ... + 5^2 - 1^2 = 101 + 97 + 93 + 89 + ... + 5 + 1 = 1326.

 

(b) (a + (2n + 1)d)^2 - (a + (2n)d)^2 + ... + (a + d)^2 - a^2 = (a + (2n + 1) d + (a + 2nd + ... + a + d) + a = n(3a + (n + 2)d).

:)

26 авг. 2020 г.
 #1
avatar+1094 
+1

Answered here by CalculatorUser

 

:)

26 авг. 2020 г.
 #2
avatar+1094 
+1

Or, instead of expanding, which could be very messy, and probably best done with an online calculator, we can use the binomial theorem.   <------ a link to what that is

 

Anways, assuming you now understand what the binomial theorem is, we can solve the problem. 
We see that the 5th power in an expansion is: (n C n-5) * x^5 * y ^[n-5]

 

In this case: 

n = 12

y = 1

x = 2x (which is weird, but the second 'x' is a value, while the first is a placeholder)

 

Plug those values in:

(12 C 7) * (2x)^5 * 1^7 =

792 * 32x^5 =

25344x^5


:)
 

26 авг. 2020 г.
 #5
avatar+1094 
+1
26 авг. 2020 г.