random problem (1 Viewer)

[damo676767]

what would this simplify too


the sum of 3(t-1)(t-2)

fron t = 1 to t = n-1

n is just a constant,

any help greatly appreaciated

 

[menty]

Water is being pumped into a pool at 2metres cubed per minute. The pool is 20m long and 5m wide. Itts depth decreases uniformly from 5m to 1m at the other end.
A) show the volume of water at depth h is 12.5h^2. Hence find he rate of increase of i) depth and ii) SA of the water when the water depth at the deepest end is on metre

 

[Stefano]

what would this simplify too


the sum of 3(t-1)(t-2)

fron t = 1 to t = n-1

n is just a constant,

any help greatly appreaciated

I don't know if I know how to do this... I'm just going to have a stab in the dark, seing as no-one else has replied.

Random answer:

S_n = 0 + 0 + 6 + 18 + 36 + 60 + 90 + ... + 3(n-2)(n-3)

GP; S<sub>n</sup> = 6[(1-6^n-1)]/-5

 

[robbo_145]

possibly trapezoidal rule but im not sure how you could get rid of the triangle at the top of each

 

[damo676767]

thanks for tring but that doesn't work

it is not a geometric or a aritmatic progrretion unfortunalty

 

[Slidey]

Stefano: that isn't a gp.

the sum of 3(t-1)(t-2)

fron t = 1 to t = n-1

3(t^2-3t+2)
The sum=
[n(n+1)(n+2)/6 + 3n(n+1)/2 + 2n]*3
=n(n+1)[n+2+9]/2 + 6n
=n(n^2+12n+9+12)/2
=n(n^2+6n+21)/2

n-1 terms, so replace n with n-1:

(n-1)(n^2-2n+1+6n-6+21)/2
=(n-1)(n^2+4n+16)/2

 

[Stefano]

Stefano: that isn't a gp.

the sum of 3(t-1)(t-2)

fron t = 1 to t = n-1

3(t^2-3t+2)
The sum=
[n(n+1)(n+2)/6 + 3n(n+1)/2 + 2n]*3
=n(n+1)[n+2+9]/2 + 6n
=n(n^2+12n+9+12)/2
=n(n^2+6n+21)/2

n-1 terms, so replace n with n-1:

(n-1)(n^2-2n+1+6n-6+21)/2
=(n-1)(n^2+4n+16)/2

Agreed, it is not a GP. But, how then, did you find the Sum Slidey? When you state that 'the sum=...' I don't follow...

 

[damo676767]

Stefano: that isn't a gp.

the sum of 3(t-1)(t-2)

fron t = 1 to t = n-1

3(t^2-3t+2)
The sum=
[n(n+1)(n+2)/6 + 3n(n+1)/2 + 2n]*3
=n(n+1)[n+2+9]/2 + 6n
=n(n^2+12n+9+12)/2
=n(n^2+6n+21)/2

n-1 terms, so replace n with n-1:

(n-1)(n^2-2n+1+6n-6+21)/2
=(n-1)(n^2+4n+16)/2

could you please explain what you did

 

[Abtari]

sliderule basically expanded the brackets thing...
then he got the expressoin

3 (t^2 - 3t + 2)

and then splitting those up, you can get sum of all the t^2 's and the sum of all the t's and sum of all the 2's.... and so forth

butttt u need to know the formula for (or be able to derive):

1^2 + 2^2 + 3^2+...+ n^2 = n(n+1)(n+2)/6
1+2+3+...+n = n(n+1)/2

otherwise you cant do it

P.S. those formulae are for sums from 1 to n, hence the need to replace at the end , 'n' with 'n-1'...thanks sliderule

 

[Stefano]

Tricky! :O

 

[damo676767]

1^2 + 2^2 + 3^2+...+ n^2 = n(n+1)(n+2)/6

i dont thin that this is true

1² + 2² + 3² +4² + 5² + 6² + 7² = 140

7*8*9/6 = 84

 

[Estel]

x^3 - (x-1)^3 = 3x^2 -3x + 1
x^2 = (1/3)[x^3 - (x-1)^3 ] + x - 1/3

so 1^2 + 2^2... + n^2
= (1/3)n^3 + n(n+1)/2 - n/3
= (2n^3 + 3n^2 + n)/6
= n(2n+1)(n+1)/6

Indeed, the formula given was incorrect.