Let S be the sum,
S = 2 + 4 + 6 + ... + 2 (n - 2) + 2 (n - 1) + 2n
Reverse the order of terms:
S = 2n + 2 (n - 1) + 2 (n - 2) + ... + 6 + 4 + 2
Add up terms in the same positions, so that twice the sum is
2S = (2n + 2) + (2n + 2) + (2n + 2) + ... + (2n + 2)
or
2S = n (2n + 2)
Divide both sides by 2 to solve for S :
S = n (n + 1)
Answer:
<u><em>−125a^(11)</em></u>
Step-by-step explanation:
<u><em>(−5a^(2)3a^(5) Then you solve</em></u>
<u><em>=−125a^(11)</em></u>
Answer:
it would be 17 im positive
Step-by-step explanation:
.
Answer:
The domain and range remain the same.
Step-by-step explanation:
Hi there!
First, we must determine what increasing <em>a</em> by 2 really does to the exponential function.
In f(x)=ab^x, <em>a</em> represents the initial value (y-intercept) of the function while <em>b</em> represents the common ratio for each consecutive value of f(x).
Increasing <em>a</em> by 2 means moving the y-intercept of the function up by 2. If the original function contained the point (0,x), the new function would contain the point (0,x+2).
The domain remains the same; it is still the set of all real x-values. This is true for any exponential function, regardless of any transformations.
The range remains the same as well; for the original function, it would have been
. Because increasing <em>a</em> by 2 does not move the entire function up or down, the range remains the same.
I hope this helps!