If a(n+1)=a(n)+.5n with a(0)=2then ....Find a(1) a(2) ect. I am not sure what a need to do to solve this problem....in order for me to find a(1) should I say...

a(1+1)=a(1+1)+.5(1)...

You wouldn't say this because the a to the left of the = sign has argument n + 1, whereas the a to the right has argument n. The two expressions won't have the same argument, since n + 1 is different than n. See more below.

and then try to solve? I dont understand the difference in a(n) and n .....what is the difference? Is it 1+1 and 1?..Thanks.

a(n) means a as a function of n, just like f(x) means f as a function of x.

You are given a(0), meaning the value of the a function at n = 0.

If you let n = 0 you get

a(0 + 1) = a(0) + .5 * 0 = 2 + 0 = 2, so since 0 + 1 = 1 we have a(1) = 2.

Then letting n = 1 you get

a(1 + 1) = a(1) + .5 * 1 = 2 + .5 = 2.5, so since 1 + 1 = 2 we have a(2) = 2.5.

Then letting n = 2 you get

a(2 + 1) = a(2) + .5 * 2 = 2.5 + 1 = 3.5, so since 2 + 1 = 3 we have a(3) = 3.5.

Then letting n = 3 you get

a(3 + 1) = a(3) + .5 * 3 = 3.5 + 1.5 = 5, so since 3 + 1 = 4 we have a(4) = 5.

You continue the process as long as necessary.

If a 50ft. whale weighs 35 tons, how much would a 60ft. whale weigh? I dont really know what kind of formaula to use to answer this...could I not just say 35/50=x/60 and then solve for x? or do I have to use the y=kx^p?? Thanks.


Why we have to use proportionality of the cube

The weights of 3-dimensional object are distributed through their volumes.

Volume can be thought of as being occupied by tiny cubes.

Two geometrically similar 3-dimensional objects can be thought of as being occupied by the same number tiny cubes. The cubes making up the larger object are just bigger.

If one object is longer than the other, then its cubes will be longer also. Of course if the cubes just got longer and
didn't get wider and higher, they wouldn't remain cubes, just like a whale that's longer but not wider and thicker wouldn't
be geometrically similar to the original whale.

So when the cubes get longer, they also get wider and higher.

In this case the larger whale is 60/50 = 1.20 times as long as the first, so its little cubes would have to be 1.2 times
longer than those of the first. That alone would give them 1.2 times the volume of the first, but we're not done because the
cubes would then have to get 1.2 times wider, and which would give us 1.2 times 1.2 times the volume of the first. Of course
we're still not done because the cubes have to get 1.2 times higher, giving us 1.2 times 1.2 times 1.2 times the volume of
the first.

This is how we end up with 1.2^3 times the volume of the first, not just 1.2 times.


Solving by Ratios

The ratio form of the solution would be of the form (W2 / W1) = ( L2 / L1)^3. You have to use the weight ratio on one side and the length ratio on the other, and should always do this when using proportionality. If you use x instead of W2 for the unknown weight of the second whale your equation would read

(x / 35) = (60 / 50)^3, or

x / 35 = 1.2^3 or

x / 35 = 1.73 so that

x = 35 * 1.73 = 61, meaning that the second whale weighs 61 tons.


Solving by Functional Proportionalies

All this is equivalent to saying that the volumes of geometrically similar objects are proportional to the cube of their lengths, or using V for volume and L for length, V = k * L^3.

The weight of a whale is distributed through its volume, so we also know that W = k * L^3, where W is the weight.

The 50-ft whale weights 35 tons so any geometrically similar whale will obey the same proportionality as this one. The proportionality is

W = k * L^3, which gives us
35 tons = k * (50 ft)^3 so that
k = 35 tons / (50 ft)^3 = 35 tons / (12500 ft^3) = .00028 tons / ft^3, approximately.

This tells us that the proportionality is

W = .00028 tons / ft^3 * L^3.

So the 60-foot whale, with L = 60 ft, would have weight

W = .00028 tons / ft^3 * (60 ft)^3 = .00028 tons / ft^3 * 216,000 ft^3 = 61 tons, approximately.

So the ratio solution gives us the same thing as the proportionality solution. </h3>