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3 years ago

Prove that one of every three consecutive positive integer is divisible by 3

1 answer:

5 0

Hello :
all n in N ; n(n+1)(n+2) = 3a a in N or : ≡ 0 (mod 3)
1 ) n ≡ 0 ( mod 3)...(1)

n+1 ≡ 1 ( mod 3)...(2)
 n+2 ≡ 2 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 0×1×2 ( mod 3) : ≡ 0 (mod 3)
2) n ≡ 1 ( mod 3)...(1)

n+1 ≡ 2 ( mod 3)...(2)
 n+2 ≡ 3 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 1×2 × 3 ( mod 3) : ≡ 0 (mod 3) , 6≡ 0 (mod)
3) n ≡ 2 ( mod 3)...(1)

n+1 ≡ 3 ( mod 3)...(2)
 n+2 ≡ 4 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 2×3 × 4 ( mod 3) : ≡ 0 (mod 3) , 24≡ 0 (mod3)

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On a certain marathon course a runner reaches a big hill that is at least 10 miles into the race. If a total marathon is 26.2 mi

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If x is the number of miles to g then

x <= 26 .2 - 10

x <= 16.2

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3 years ago

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The graphs show how many words per minute 2 students read

marusya05 [52]

Answer:

Step-by-step explanation:

first student read 120 words words per minute.

second student read 185 words per minute.

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3 years ago

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The blood pressure in millimeters was measured for a large sample of people. The average pressure is 140 mm, and the SD of the m

Amanda [17]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Assum a normal distribution :

Mean (m) = 140 mm

Standard deviation (sd) = 20 mm

The percentage of people with blood pressure between 115 and 165 mm.

Zscore = (x - m) / sd

X = 115 mm

(115 - 140) / 20

-25/20 = - 0.625 = - 0.63

P(z<-0.63) = 0.2643

X = 165 mm

(165 - 140) / 20

25/20 = 0.625 = 0.63

P(z< 0.63) = 0.7357

0.7357 - 0.2643 = 0.4714 = 47.14%

B.) The percentage of people with blood pressure between 140 and 165 mm.

Zscore = (x - m) / sd

X = 140 mm

(140 - 140) / 20

0/20 = 0 = 0

P(z<0) = 0.5000

X = 165 mm

(165 - 140) / 20

25/20 = 0.625 = 0.63

P(z< 0.63) = 0.7357

0.7357 - 0.5000 = 0.2357 = 23.57%

C.) ___ The percentage of people with blood pressure over 165 mm.

X = 165mm

(165 - 140) / 20

25/20 = 0.625 = 0.63

1 - P(z< 0.63) = 0.7357

1 - 0.7357 = 0.2643 * 100% = 26.43%

8 0

3 years ago

Evaluate the infinite sum:

satela [25.4K]

Consider the k-th partial sum,

$S_k = 1 + \dfrac2\pi + \dfrac3{\pi^2} + \cdots + \dfrac k{\pi^{k-1}}$

More compactly,

$\displaystyle S_k = \sum_{i=1}^k \frac i{\pi^{i-1}} = \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}}$

(this is just another case of a similar sum you asked about a while ago [24494877])

The infinite sum is the limit of the partial sum as k goes to infinity. We have

$\displaystyle \lim_{k\to\infty} \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}} = \frac\pi{(1-\pi)^2} \lim_{k\to\infty} \left(\frac{(1-\pi)k}{\pi^k} + \pi - \frac1{\pi^{k-1}} \right) = \boxed{\frac{\pi^2}{(1-\pi)^2}}$

since the non-constant terms in the limit converge to 0.

Alternatively, recall that for |x| < 1, we have

$\dfrac1{1-x} = \displaystyle \sum_{n=0}^\infty x^n$

Differentiating both sides gives

$\dfrac1{(1-x)^2} = \displaystyle \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}$

also valid for |x| < 1. Take x = 1/π and you get the sum you want to compute.

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3 years ago

What steps would you do to solve n/7+3 greater than or equal to -4?

user100 [1]

Answer:

the answer is n ≥ -49

here are the steps for solving-

the problem Is like n/7+3 ≥ -4

1. First subtract 3 from both sides. After doing that, the inequation should be Like This:

n/7 ≥ -7

2. since 7 is in Division form, when it goes to RHS, You multiply.

So -7 times 7 Is -49

so the answer is n≥ -49

5 0

3 years ago