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Sequence Question

  • Thread starter l46kok
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  • #1
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Homework Statement


U( r ) = r(r+1)(r+2)(r+3), show that U (r + 1) - U ( r ) = 4(r+1)(r+2)(r+3) Hence, find the sum to n terms of the series 2x3x4 + 3x4x5 + 4x5x6 +.....



Homework Equations



Sequence Knowledge..

The Attempt at a Solution



I can't even TOUCH on the problem. Can you please help?
 

Answers and Replies

  • #2
tiny-tim
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Hi l46kok! :smile:
U( r ) = r(r+1)(r+2)(r+3), show that U (r + 1) - U ( r ) = 4(r+1)(r+2)(r+3)
I assume you can do that?
Hence, find the sum to n terms of the series 2x3x4 + 3x4x5 + 4x5x6 +.....
ok, define U(r + 1) - U( r ) = V(r).

Then the question is asking you for 1/4 ∑ V(r).

Hint: what is V(r) + V(r+1), in terms of Us ? :wink:
 
  • #3
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Hi l46kok! :smile:


I assume you can do that?


ok, define U(r + 1) - U( r ) = V(r).

Then the question is asking you for 1/4 ∑ V(r).

Hint: what is V(r) + V(r+1), in terms of Us ? :wink:
U( r ) = r(r+1)(r+2)(r+3), show that U (r + 1) - U ( r ) = 4(r+1)(r+2)(r+3)

U(r+1) is = (r+1)(r+2)(r+3)(r+4)

so U(r+1) - U( r ) = (r+1)(r+2)(r+3)(r+4) - r(r+1)(r+2)(r+3)

let (r+1)(r+2)(r+3) = z

= z((r+4)-r)

= 4z

= 4(r+1)(r+2)(r+3)



Then if we say V(r) = U(r+1) - U( r )

V(r) + V(r + 1) = U(r+1) - U(r) + U(r+2) - U(r+1)

= U(r+2) - U(r)

Then we're finding 1/4 of Summation of V(r) so

Summation of 0.25V(r) = 0.25 U(r+n) - U(r)

?? is this it?

oh yeah and where did the 1/4 come from?
 
Last edited:
  • #4
HallsofIvy
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U( r ) = r(r+1)(r+2)(r+3), show that U (r + 1) - U ( r ) = 4(r+1)(r+2)(r+3)

U(r+1) is = (r+1)(r+2)(r+3)(r+4)

so U(r+1) - U( r ) = (r+1)(r+2)(r+3)(r+4) - r(r+1)(r+2)(r+3)

let (r+1)(r+2)(r+3) = z

= z((r+4)-r)

= 4z

= 4(r+1)(r+2)(r+3)
Excellent!



Then if we say V(r) = U(r+1) - U( r )

V(r) + V(r + 1) = U(r+1) - U(r) + U(r+2) - U(r+1)

= U(r+2) - U(r)

Then we're finding 1/4 of Summation of V(r) so

Summation of 0.25V(r) = 0.25 U(r+n) - U(r)

?? is this it?

oh yeah and where did the 1/4 come from?
Since V(r)= U(r+1)- U(r), V(1)+ v(2)= U(2)- U(1)+ U(3)- U(2)= U(3)- U(1).

V(1)+ V(2)+ V(3)= U(2)- U(1)+ U(3)- U(2)+ U(4)- U(3)= U(4)- U(1).

V(1)+ V(2)+ V(3)+ V(4)= U(2)- U(1)+ U(3)- U(2)+ U(4)- U(3)+ U(5)- U(4)= U(5)- U(1).

Get the point? This is a "telescoping series" [itex]\sum_{i=1}^n V(i)= U(n+1)- U(1)[/iyrc].

The "1/4" is to get rid of the "4" in the formula for V(r). Since V(r)= 4(r+1)(r+2)(r+3), the sum you are asked to do is (1/4)(V(1)+ V(2)+ ...+ V(n)).
 

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