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harkovskaia [24]

3 years ago

12

Write a for loop that computes the following sum: 5 10 15 20 ... 485 490 495 500. The sum should be placed in a variable sum tha

t has already been declared and initialized to 0. In addition, there is another variable, num that has also been declared. You must not use any other variables.

1 answer:

Varvara68 [4.7K]3 years ago

8 0

The solution is in the attachment

You might be interested in

Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]

Schach [20]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

$\rm :\longmapsto\:n + {n}^{2} + {n}^{3} + - - - + {n}^{n}$

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

$\rm \: = \: \dfrac{n( {n}^{n} - 1)}{n - 1}$

$\rm \: = \: \dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }$

Now, Consider Denominator, we have

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n}$

can be rewritten as

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {(n - 1)}^{n} + {n}^{n}$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

So, on substituting the values evaluated above, we get

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

Now, we know that,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}$

So, using this, we get

$\rm \: = \: \dfrac{1}{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

$\rm \: = \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{e}{e - 1} }$

$\rm \: = \: \dfrac{e - 1}{e}$

$\rm \: = \: 1 - \dfrac{1}{e}$

Hence,

$\boxed{\tt{ \displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} } = \frac{e - 1}{e} = 1 - \frac{1}{e}}}$

3 0

2 years ago

. The number three thousand two hundred fifty-one written in standard form would be 3,251. True False

OLga [1]

This would be true.
All the place values and it numbers are correct.

Your answer would be true.

Hope this helps!

4 0

2 years ago

Read 2 more answers

Sean wrote the equation 9.25 = 5m, where m is the cost per pound, to show the relationship between the total cost, $9.25, and th

OLEGan [10]

<h2>Explanation:</h2>

The complete question is in a comment above. The statements says that:

Sean wrote the equation $9.25=5m$
$m$ is the cost per pound.
The equation show the relationships between the total cost, $9.25, and the number of pounds of pears.

So we are asked to write an write an equation to show the cost y of x pounds of pears. In other words, we need to write an equation such that y is given in terms of x, where x is the pounds of pears. So:

$y=f(x)$

From the equation we can determine $m$ as:

$m=\frac{9.25}{5}=1.85$

So each pound costs $1.85. The equation tells us that 5 pounds of pears result costs $9.25. Then, by proportion:

$\frac{y}{x}=\frac{9.25}{5} \\ \\ y=1.85x$

So choosing the number of pounds as 0, 1, 2, 3, 4, 5, the table would be:

$\begin{array}{cc}x(\text{Number of pounds}) & y(\text{Cost in dollars})\\0 & 0\\1 & 1.85\\2 & 3.7\\3 & 5.55\\4 & 7.4\\5 & 9.25\end{array}$

From the table we can know that the graph is a line. that passes through all these points. Since this line passes through the origin, then y varies directly as x (Direct Proportion).

3 0

3 years ago

What is the slope of the line through (-7,-8) and (0,4)

Oksanka [162]

Answer:

c.12/7

Step-by-step explanation:

3 0

2 years ago

Max has a drawer full of green and red socks. He has three times times as many red socks subtracted from four times as many gree

monitta

Answer:

There are 42 red colour socks and 44 green color socks

Step-by-step explanation:

Let there are r red socks and g green socks.

ATQ,

He has three times times as many red socks subtracted from four times as many green socks which he believes is 50 socks.

4g-3r=50 ....(1)

Half the number of green socks plus one-third of the number of red socks is 36.

$\dfrac{g}{2}+\dfrac{r}{3}=36\\\\3g+2r=216\ .....(2)$

Multiply equation (1) by 2 and equation (2) by 3.

8g-6r = 100 ....(3)

9g +6r = 648 ....(4)

Add equation (3) and (4)

8g-6r + 9g +6r = 100+648

17g = 748

g = 44

Put the value of g in equation (1).

4(44)-3r=50

176-3r = 50

176-50 = 3r

r = 42

Hence, there are 42 red colour socks and 44 green color socks.

7 0

2 years ago