Answer:
a. f(n) = n(n+1)/2
b. 351
c. 19
d. no; n would not be an integer for f(n) = 17.
Step-by-step explanation:
a. The terms are 1, 3, 6, 10, ...
Each term has the term number added to the previous term. That is, first differences are 3-1 = 2, 6-3 = 3, 10-6 = 4. These differences have a constant difference of 1.
When the 2nd differences are constant, a 2nd degree polynomial can describe the sequence. It is a little bit of trouble to find that polynomial.
For the polynomial ...
[tex]f(n)=an^2+bn+c[/tex]
We can substitute values for n and f(n) to get 3 equations in 3 unknowns:
[tex]1=a\cdot1^2+b\cdot1+c\\\\3=a\cdot2^2+b\cdot2+c\\\\6=a\cdot3^2+b\cdot3+c[/tex]
Solving these equations by your favorite method gives ...
(a, b, c) = (1/2, 1/2, 0)
That is, the function representing the relationship is ...
f(n) = n(n+1)/2 . . . . the function describing the relationship
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b. f(26) = 26(27) = 13(27) = 351 . . . . the number of squares in term 26
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c. The number of the term having 190 squares can be found by solving ...
n(n +1)/2 = 190
n(n +1) = 380 = 19(20)
The 19th term will have 190 squares.
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d. Terms 5 and 6 are 15 and 21.
17 is not the value of one of the terms in this sequence.
