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Could the triangle have 140 cans? No.
a= 2, d= 1
Identity Property of Multiplication
Remove parentheses
Subtract term
The explicit formula for the sequence of the number of cans is a_n=n+1.
∑_(n= 1)^?(n+1)
The triangle has 9 rows. This is the upper limit for the summation. ∑_(n=1)^9(n+1)
There are 18 cans in the 17th row.
S_n=n/2(a_1 + a_n)
Here, n is the number of terms of the series and a_1 and a_n are the first and last terms, respectively. We know that the first term of our series is a_1= 2. Moreover, from Part A, we know that a_n=n+1.
S_n= 110
LHS * 2=RHS* 2
Add terms
Distribute n
LHS-220=RHS-220
Commutative Property of Addition
Rearrange equation
Note that we have a quadratic equation with coefficients a= 1, b= 3, and c= - 220. To solve it we will use the Quadratic Formula.
Substitute values
Calculate power
Identity Property of Multiplication
a(- b)=- a * b
a-(- b)=a+b
Use a calculator
State solutions
(I), (II): Add and subtract terms
(I), (II): Calculate quotient
Be aware that n represents the nth row in the triangle. Therefore, a value of n that is not a natural number does not make sense. We supposed the triangle had 110 cans, and arrived to the conclusion that there are either 13.4 or - 16.4 rows. None of these numbers make sense, so the triangle cannot have 110 cans. By the same method let's see if the triangle can have 140 cans.
| Cans | Substitute | Simplify | Value of n |
|---|---|---|---|
| 110 | 110 = n/2(2+n+1) | n^2+3n-220=0 | n≈ 13.4 or n≈ - 16.4 |
| 140 | 140 = n/2(2+n+1) | n^2+3n-280=0 | n≈ 15.3 or n≈ - 18.3 |
Since the triangle cannot have 15.3 nor - 18.3 rows, we conclude that it cannot have 140 cans.