Mathematics, 07.04.2020 16:24 asclp

EXAMPLE 2 Find a formula for the general term of the sequence 3 5 , − 4 25 , 5 125 , − 6 625 , 7 3125 , assuming that the pattern of the first few terms continues. SOLUTION We are given that a1 = 3 5 a2 = − 4 25 a3 = 5 125 a4 = − 6 625 a5 = 7 3125 . Notice that the numerators of these fractions start with 3 and increase by 1 Correct: Your answer is correct. whenever we go to the next term. The second term has numerator 4, the third term has numerator 5; in general, the nth term will have numerator Incorrect: Your answer is incorrect. . The denominators are powers of 5 Correct: Your answer is correct. , so an has denominator Correct: Your answer is correct. . The signs of the terms are alternately positive and negative so we need to multiply by a power of −1. Here we want to start with a positive term and so we use (−1)n − 1 or (−1)n + 1. Therefore, an = (−1)n − 1 · Incorrect: Your answer is incorrect. .

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Answer from: ander67061

The given terms can be written in the generalized format as

$\rm a_n=\dfrac{(2n+1)(-1)^{n+1}}{5^n}$ and this can be done by observing the given terms.

Given :

General Terms --

$\rm a_1 = \dfrac{3}{5}$ $\rm a_2 = -\dfrac{4}{25}$ $\rm a_3 = \dfrac{5}{125}$ $\rm a_4 = -\dfrac{6}{625}$ $\rm a_5 = \dfrac{7}{3125}$

The term $\rm a_2$ can be written as:

$\rm a_2 = \dfrac{4}{5^2}$

The term $\rm a_3$ can be written as:

$\rm a_3 =\dfrac{5}{5^3}$

The term $\rm a_4$ can be written as:

$\rm a_4 =\dfrac{6}{5^4}$

The term $\rm a_5$ can be written as:

$\rm a_5 =\dfrac{7}{5^5}$

So, the above terms can be written in the generalized format as:

$\rm a_n=\dfrac{(2n+1)(-1)^{n+1}}{5^n}$

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Answer from: hey743358

$a_n=\dfrac{(2+n)\cdot(-1)^{n + 1}}{5^n}$

Step-by-step explanation:

We are to find a formula for the general term of the sequence

$\dfrac{3}{5}, -\dfrac{4}{25}, \dfrac{5}{125}, -\dfrac{6}{625}, \dfrac{7}{3125}$

We are given that $a_1=\dfrac{3}{5}, a_2=-\dfrac{4}{25}, a_3=\dfrac{5}{125}, a_4=-\dfrac{6}{625}, a_5=\dfrac{7}{3125}$ .

The numerators of these fractions start with 3 and increase by 1. The second term has numerator 4, the third term has numerator 5; in general, the nth term will have numerator 2+nThe denominators are powers of 5, so a_n

has denominator 5^n

The signs of the terms are alternately positive and negative so we need to multiply by a power of −1. Here we want to start with a positive term and so we use $(-1)^{n + 1}$ .

Therefore,

$a_n=\dfrac{(2+n)\cdot(-1)^{n + 1}}{5^n}$

Answer from: Quest

step-by-step explanation:

Answer from: Quest

answer:

total cost for tiles and paints is $924.

step-by-step explanation:

we have been given that a community hall is in the shape of a cuboid. the hall is 40m long 15m high and 3m wide.

the paint will be required for 4 walls and ceiling.

let us find area of walls and ceiling.

$\text{area of walls and ceiling}=(2*40*15)+(2*3*15)+(40*3)$

$\text{area of walls and ceiling}=1200+90+120$

$\text{area of walls and ceiling}=1410$

therefore, the area of walls and ceiling is 1410 square meters.

given: cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. therefore,

$\text{ the total painting cost}=10*(\frac{1410}{25})$

$\text{ the total painting cost}=10*56.4=564$

therefore, the total painting cost is $564.

tiles will be required for floor. let us find the area of floor.

$\text{area of floor} = 40*3\text{ square meters}$

$\text{area of floor} =120\text{ square meters}$

given: 1m squared floor tiles costs $3. so,

$\text{total cost for tiles} = 3*120 = 360$

therefore, the total tiles cost is $360.

now let us find combined total cost of tiles and paint.

$\text{combined total cost}= 564+360 = 924$

therefore, the combined total cost of tiles and paint is $924.

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EXAMPLE 2 Find a formula for the general term of the sequence 3 5 , − 4 25 , 5 125 , − 6 625 , 7 312...

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