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1 year ago

Prove usin's Mathematical induction 2^n ≤(n+1)!, for n≥0

Mathematics

1 answer:

Westkost [7]1 year ago

8 0

Answer:

Step-by-step explanation:

2^0 is less than or equal to 1!, because 1<= 1

if 2^n <= (n+1)!, we wish to show that 2^(n+1) <= (n+2)!, since

(n+2)! = (n+1)! * (n+2), and (n+1)!>= 2^n, then we want to prove that n+2<=2, which is always true for n>=0

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Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

2 years ago

Which question represents the line through (0,0) and (2,5)?

madam [21]

Answer:

y=5/2x

Step-by-step explanation:

The line needs to go through the points (0,0) and (2,5).

With y=5/2x, there is no y-intercept, so it starts on the origin, (0,0). With the slope 5/2, you would rise 5 (y) from the origin, and run 2 (x), which gives you the point (2,5).

4 0

1 year ago

Multiply and simplify if possible

Natalka [10]

Answer:

Step-by-step explanation:

i think its correct

6 0

2 years ago

2x + 6y = -6

loris [4]

Answer:

(-3, 0)

Step-by-step explanation:

Solve by elimination. Multiply the first equation by -2 and then add the two equations together.

6 0

2 years ago

Read 2 more answers

All cheeses in Blake's Grocery sell for no less than $3.95 per pound

Greeley [361]

Answer:

3.95,3.96,3.97,3.98,3.99 i think lol

Step-by-step explanation:

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2 years ago

Prove usin's Mathematical induction 2^n ≤(n+1)!, for n≥0​

Prove usin's Mathematical induction 2^n ≤(n+1)!, for n≥0