Answer:
Step-by-step explanation:
The first differences of the sequence are ...
- 5-2 = 3
- 10-5 = 5
- 17-10 = 7
- 26-17 = 9
- 37-26 = 11
Second differences are ...
- 5 -3 = 2
- 7 -5 = 2
- 9 -7 = 2
- 11 -9 = 2
The second differences are constant, so the sequence can be described by a second-degree polynomial.
We can write and solve three equations for the coefficients of the polynomial. Let's define the polynomial for the sequence as ...
f(n) = an^2 + bn + c
Then the first three terms of the sequence are ...
- f(1) = 2 = a·1^2 + b·1 + c
- f(2) = 5 = a·2^2 +b·2 + c
- f(3) = 10 = a·3^2 +b·3 +c
Subtracting the first equation from the other two gives ...
3a +b = 3
8a +2b = 8
Subtracting the first of these from half the second gives ...
(4a +b) -(3a +b) = (4) -(3)
a = 1 . . . . . simplify
Substituting into the first of the 2-term equations, we get ...
3·1 +b = 3
b = 0
And substituting the values for a and b into the equation for f(1), we have ...
1·1 + 0 + c = 2
c = 1
So, the formula for the sequence is ...
f(n) = n^2 + 1
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The 20th term is f(20):
f(20) = 20^2 +1 = 401
_____
<em>Comment on the solution</em>
It looks like this matches the solution of the "worked example" on your problem page.
Answer:
C
Step-by-step explanation:
(2x + 3)^5 = C(5,0)2x^5*3^0 +
C(5,1)2x^4*3^1 + C(5,2)2x^3*3^2 + C(5,3)2x^2*3^3 + C(5,4)2x^1*3^4 + C(5,5)2x^0*3^5
Recall that
C(n,r) = n! / (n-r)! r!
C(5,0) = 1
C(5,1) = 5
C(5,2) = 10
C(5,3) = 10
C(5,4) = 5
C(5,5) = 1
= 1(2x^5)1 + 5(2x^4)3 + 10(2x^3)3^2 + 10(2x^2)3^3 + 5(2x^1)3^4 + 1(2x^0)3^5
= 2x^5 + 15(2x^4) + 90(2x^3) + 270(2x^2) + 405(2x) +243
= 32x^5 + 15(16x^4) + 90(8x^3) + 270(4x^2) + 810x + 243
= 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243
Step-by-step explanation:
- 1.25 × 10^-4
- 2.0 × 10^-4
- 3.25×10^-6
Answer:
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Step-by-step explanation:
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The greatest common factors of 30,48,60 is 6.
