This the sequence.
nth term 3,4,5,6,7
no. of diagonals 0,2,5,9,14
A polygon with n sides has a total of
1/p * n(n-q) diagonals, where p and q are integers.
Find the values of p and q.
Another question (related to the previous one) -
A polygon with n + 1 sides has 30 more diagonals than a polygon with n sides.
Find n.
nth term 3,4,5,6,7
no. of diagonals 0,2,5,9,14
A polygon with n sides has a total of
1/p * n(n-q) diagonals, where p and q are integers.
Find the values of p and q.
Another question (related to the previous one) -
A polygon with n + 1 sides has 30 more diagonals than a polygon with n sides.
Find n.
-
A polygon with n sides has a total of (1/2) * n * (n - 3) sides, so
p = 2
q = 3
You can find them by inspection, the numbers are pretty small. If you want a more intuitive answer, you can think about the problem combinatorially (i.e., as a counting problem). It basically boils down to asking how many ways you can choose two of the corners on a shape to draw a line between, but this is a slight overestimate since you might choose one of the edges of the shape as your two points. A simple case to picture is the square. You could pick any two of the four corners and draw a line between them. However, if you picked two corners on the same side, and drew a line, you wouldn't make a diagonal. Hence, the formula is (n choose 2) - n, where n is the number of sides. This formula simplifies using the definition of the binomial coefficient. There is a further discussion on proving that this result is true by induction at the link given in the source.
Knowing this, you can solve the next problem by dropping in n + 1 for n in the equation, then setting up the equality and simplifying:
(1/2) * (n + 1) * (n + 1 - 3) = (1/2) * n * (n - 3) + 30
(1/2)(n^2 - n) - 1 = (1/2)(n^2 - 3n) + 30
n = 31
p = 2
q = 3
You can find them by inspection, the numbers are pretty small. If you want a more intuitive answer, you can think about the problem combinatorially (i.e., as a counting problem). It basically boils down to asking how many ways you can choose two of the corners on a shape to draw a line between, but this is a slight overestimate since you might choose one of the edges of the shape as your two points. A simple case to picture is the square. You could pick any two of the four corners and draw a line between them. However, if you picked two corners on the same side, and drew a line, you wouldn't make a diagonal. Hence, the formula is (n choose 2) - n, where n is the number of sides. This formula simplifies using the definition of the binomial coefficient. There is a further discussion on proving that this result is true by induction at the link given in the source.
Knowing this, you can solve the next problem by dropping in n + 1 for n in the equation, then setting up the equality and simplifying:
(1/2) * (n + 1) * (n + 1 - 3) = (1/2) * n * (n - 3) + 30
(1/2)(n^2 - n) - 1 = (1/2)(n^2 - 3n) + 30
n = 31
-
This the sequence.
nth term 3,4,5,6,7 ..................... Tn = n + 2
no. of diagonals 0,2,5,9,14
A polygon with n sides has a total of C(n, 2) -- n diagonals = n(n -- 3) / 2
1/p * n(n-q) diagonals, where p = 2 and q = 3.
A polygon with n + 1 sides has 30 more diagonals than a polygon with n sides.
whence (n + 1)(n -- 2) = n(n -- 3) + 60 giving n = 31
nth term 3,4,5,6,7 ..................... Tn = n + 2
no. of diagonals 0,2,5,9,14
A polygon with n sides has a total of C(n, 2) -- n diagonals = n(n -- 3) / 2
1/p * n(n-q) diagonals, where p = 2 and q = 3.
A polygon with n + 1 sides has 30 more diagonals than a polygon with n sides.
whence (n + 1)(n -- 2) = n(n -- 3) + 60 giving n = 31