Evaluate the series 9 + 3 + 1 + . . .
The given series is an arithmetic series with a common difference of −2, starting from 9.
Using the formula for the sum of an arithmetic series, we can calculate the sum Sₙ of the first n terms:
Sₙ = (n/2) * (a₁ + aₙ)
= (n/2) * (9 + aₙ)
where a₁ is the first term (9) and aₙ is the nth term.
The nth term can be determined using the formula for the nth term of an arithmetic series:
aₙ = a₁ + (n-1)d
= 9 + (n-1)(-2)
= 9 - 2n + 2
= 11 - 2n
Substituting this expression for aₙ into the formula for Sₙ:
Sₙ = (n/2) * (9 + 11 - 2n)
= (n/2) * (20 - 2n)
= 10n - n²
Therefore, the sum of the series is represented by the equation 10n - n².
However, since the series is unbounded and does not converge, we cannot determine a finite sum.
are you sure
Apologies for the confusion in my previous response. I made a mistake in assuming the series was unbounded when it is actually bounded.
The series given is 9 + 3 + 1 + ...
We can observe that the common difference between consecutive terms is -6.
To find the sum of this arithmetic series, we can use the formula for the sum of an arithmetic series:
Sₙ = (n/2) * (a₁ + aₙ)
where Sₙ is the sum of the first n terms, a₁ is the first term, and aₙ is the nth term.
The nth term can be determined using the formula for the nth term of an arithmetic series:
aₙ = a₁ + (n-1)d
= 9 + (n-1)(-6)
= 9 - 6n + 6
= 15 - 6n
Substituting this expression for aₙ into the formula for Sₙ:
Sₙ = (n/2) * (9 + 15 - 6n)
= (n/2) * (24 - 6n)
= 12n - 3n²
Therefore, the sum of the series 9 + 3 + 1 + ... can be represented by the equation 12n - 3n².