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Dmitry [639]
3 years ago
14

Find the sum of the first 6 terms of geometric series 2-10+50+.....

Mathematics
1 answer:
barxatty [35]3 years ago
8 0
<h3>Answer:   -5208</h3>

============================================

Work Shown:

term1 = 2

term2 = -10

term3 = 50

term4 = -250

term5 = 1250

term6 = -6250

Each new term is found by multiplying the previous term by -5

Add up the terms

2 + (-10) + 50 + (-250) + 1250 + (-6250) = -5208

-------------

A shortcut is to use a = 2, r = -5 and n = 6 in the formula below

Sn = a*(1-r^n)/(1-r)

S6 = 2*(1-(-5)^6)/(1-(-5))

S6 = 2*(1-15625)/(1+5)

S6 = 2*(-15624)/6

S6 = -31248/6

S6 = -5208

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f(x) to f(x)-3 is the whole graph of f(x), shifted 3 units down. 

f(x) to -2f(x): 

The effect of "multiplication by -" is that the whole graph is reflected with respect to the x axis, so it is turned upside down.
 
The effect of "multiplication by 2" is that every point is "stretched vertically by a factor of 2" . So for example the point (-1, -4) in the original function, becomes (-1, -8) in the second one. Or (2, 5) would become (2, 10). 

The only points that do not change (are not streched vertically) are the roots. For example if (4,0) is an x-intercept (a root) in the original function, (4,0) is still a root in the second one because  2 times 0 is still 0.


2. Consider the polynomial function of degree n: 

f(x)= a_{n} x^{n} +a_{n-1} x^{n-1}+....+a_{2} x^{2}+a_{1} x^{1}+a_{0}

a. Y-intercept

The y - intercept is the value of the polynomial function at x=0. 
So it is f(0)=a_{0}, that is, the constant term of f(x)

in f(x)-3 the y intercept is shifted 3 units down as any other point, so it becomes  a_{0}-3

In -2f(x), the y-intercept a_{0} becomes -2a_{0}

b. Regions of f decreasing or increasing

f(x)-3 is f(x) just shifted down 3 units, so they are both increasing and decreasing in the same intervals of x

-2f(x) is f(x) turned upside down, so -2f(x) is increasing in all intervals f(x) is decreasing and it is decreasing in all intervals f(x) is increasing.

c. End behaviors

By now it is clear that end behaviors of f(x) and f(x)-3 are same, and f(x) with -2f(x) are opposite

d. Evenness, oddness

If f(x) is even, then f(x)=f(-x)

Let g(x)=f(x)-3

g(x)=f(x)-3=f(-x)-3=g(-3), so in this case f(x)-3 is even

If f(x) is odd, then f(-x)=-f(x)

g(x)=f(x)-3=-f(-x)-3,

so -g(x)=f(-x)+3

g(-x)=f(-x)-3,  

so g(-x) is not equal to -g(x). Which means f(x)-3 is not odd if f(x) is


Consider f(x)=-2f(x)

If f(x) is even, f(x)=f(-x)

g(x)=-2f(x)=-2f(-x)
g(-x)=-2f(-x)

So g(x)=g(-x), which means -2f(x) is even if f(x) is even

If f(x) is odd, f(x)=-f(-x)

let g(x)=-2f(x)=-2(-f(-x))=2f(-x)

g(-x)=-2f(-x)=-2(-f(x))=2f(x)

so g(-x) is not equal to -g(x), thus -2f(x) is not odd if f(x) is odd.

The conclusions about oddness and evenness can be also derived from the discussions about the graphs.
 

6 0
3 years ago
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