Questions: Suppose that the sum from n=1 to infinity of an = -2 and the sum from n=1 to infinity of bn = 1 and a1 = 5 and b1 = 7, find the sum of the series: A. The sum from n=1 to infinity of (-3 an + -1 bn) = B. The sum from n=2 to infinity of (-3 an + -1 bn) =

Suppose that the sum from n=1 to infinity of an = -2 and the sum from n=1 to infinity of bn = 1 and a1 = 5 and b1 = 7, find the sum of the series:
A. The sum from n=1 to infinity of (-3 an + -1 bn) =
B. The sum from n=2 to infinity of (-3 an + -1 bn) =

Transcript text: Suppose that $\sum_{n=1}^{\infty} a_{n}=-2$ and $\sum_{n=1}^{\infty} b_{n}=1$ and $a_{1}=5$ and $b_{1}=7$, find the sum of the series: A. $\sum_{n=1}^{\infty}\left(-3 a_{n}+-1 b_{n}\right)=$ $\square$ B. $\sum_{n=2}^{\infty}\left(-3 a_{n}+-1 b_{n}\right)=$ $\square$ Submit Question

Solution

Solution Steps

To solve these problems, we need to use the properties of infinite series and linearity.

A. For the series $\sum_{n=1}^{\infty}(-3a_n - b_n)$, we can use the linearity of summation: $\sum_{n=1}^{\infty}(-3a_n - b_n) = -3\sum_{n=1}^{\infty}a_n - \sum_{n=1}^{\infty}b_n$. Substitute the given sums to find the result.

B. For the series $\sum_{n=2}^{\infty}(-3a_n - b_n)$, we need to subtract the first terms from the total sums: $\sum_{n=2}^{\infty}(-3a_n - b_n) = \sum_{n=1}^{\infty}(-3a_n - b_n) - (-3a_1 - b_1)$.

Step 1: Calculate the Sum for Part A

We start with the series $\sum_{n=1}^{\infty}(-3a_n - b_n)$. Using the linearity of summation, we have:

\[ \sum_{n=1}^{\infty}(-3a_n - b_n) = -3\sum_{n=1}^{\infty}a_n - \sum_{n=1}^{\infty}b_n \]

Substituting the given values:

\[ = -3(-2) - 1 = 6 - 1 = 5 \]

Thus, the result for part A is:

\[ \sum_{n=1}^{\infty}(-3a_n - b_n) = 5 \]

Step 2: Calculate the Sum for Part B

Next, we consider the series $\sum_{n=2}^{\infty}(-3a_n - b_n)$. We can express this as:

\[ \sum_{n=2}^{\infty}(-3a_n - b_n) = \sum_{n=1}^{\infty}(-3a_n - b_n) - (-3a_1 - b_1) \]

Substituting the previously calculated sum and the values of $a_1$ and $b_1$:

\[ = 5 - (-3(5) - 7) = 5 - (-15 - 7) = 5 - (-22) = 5 + 22 = 27 \]