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GrogVix [38]
3 years ago
12

How is the sum expressed in sigma notation? 22 + 27 + 32 + 37 + 42 + 47

Mathematics
1 answer:
adelina 88 [10]3 years ago
5 0

The sum expressed in sigma notation is \sum_{n=1}^{6} 5 n+17.

Solution:

Given series is 22 + 27 + 32 + 37 + 42 + 47.

Write it in the sequence form 22, 27, 32, 37, 42, 47.

First term of the sequence = 22

Common difference = 27 – 22 = 5

This is arithmetic sequence.

\text n^{th} term of the arithmetic sequence is

a_n=a_1+(n-1)d

\Rightarrow a_n=22+(n-1)5

\Rightarrow a_n=22+5n-5

\Rightarrow a_n=17+5n

\Rightarrow a_n=5n+17

In the given series number of terms is 6.

So n = 1 to 6.

Write n = 1 in the lower of the sigma and 6 in the top of the sigma.

\sum_{n=1}^{6} 5 n+17

The sum expressed in sigma notation is \sum_{n=1}^{6} 5 n+17.

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Use the factor theorem to determine whether or not the first polynomial is a factor of the second
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Answer/Step-by-step explanation:

Based on the factor theorem, a polynomial, x - a, is said to be a factor of another polynomial, f(x), if an only if f(a) = 0.

Using this theorem, let's determine whether each of the given first polynomial, is a factor of the second polynomial.

1. x - 1; x² + 2x + 5

By the factor theorem, x - 1 will be a factor of f(x) = x² + 2x + 5, if and only if f(1) = 0.

f(1) = 1² + 2(1) + 5

= 1 + 2 + 5

= 8

Since f(1) ≠ 0, therefore the first polynomial, x - 1, is NOT a factor of the second polynomial, x² + 2x + 5.

2. x + 1; x³ - x - 2

By the factor theorem, x + 1 will be a factor of f(x) = x³ - x - 2, if and only if f(-1) = 0.

f(1) = -1³ - (-1) - 2

= -1 + 1 - 2

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Since f(-1) ≠ 0, therefore the first polynomial, x + 1, is NOT a factor of the second polynomial, x³ - x - 2.

3. x - 4; 2x³ - 9x² + 9x - 20

By the factor theorem, x - 4 will be a factor of f(x) = 2x³ - 9x² + 9x - 20, if and only if f(4) = 0.

f(4) = 2(4)³ - 9(4)² + 9(4) - 20

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= 0

Since f(4) = 0, therefore the first polynomial, x + 4, is a factor of the second polynomial, 2x³ - 9x² + 9x - 20.

4. a - 1; a³ - 2a² + a - 2

By the factor theorem, a - 1 will be a factor of f(a) = a³ - 2a² + a - 2, if and only if f(1) = 0.

f(1) = 1³ - 2(1)² + 1 - 2

= 1 - 2 + 1 - 2

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Since f(1) ≠ 0, therefore the first polynomial, a - 1, is NOT a factor of the second polynomial, a³ - 2a² + a - 2.

5. y + 3; 2y³ + y² - 13y + 6

By the factor theorem, y + 3 will be a factor of f(y) = 2y³ + y² - 13y + 6, if and only if f(-3) = 0.

f(-3) = 2(-3)³ + (-3)² - 13(-3) + 6

= -54 + 9 + 39 + 6

= 0

Since f(-3) = 0, therefore the first polynomial, y + 3, is a factor of the second polynomial, 2y³ + y² - 13y + 6.

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