Math Is Fun Forum

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

problem doing summation

I'm having troubles doing this sum as part of a past paper im doing for exam friday.

Theres an image of a region of the graph y = 3^x between 0 and 1 with a set of strips shown beneath the curve with width 'h' so that nh = 1 where n is the number of strips

i) By using the set of rectangles indicated on the diagram show that A > 2h/(3^h - 1) where A is area of the curve between 0 and 1

Now, summing the areas of the strips, i get h(1+3^h + 3^2h .... + 3^[h(n-1)]) but im having trouble trying to get that to be 2h/(3^h - 1)

I know how to sum a geometric series from way back in the modules, but if i plug it in here i get.

Have i done something wrong, because i cannot find a way of equating that to the expression given.

Last edited by luca-deltodesco (2008-05-19 04:13:36)

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Identity

Member

Registered: 2007-04-18

Posts: 934

Re: problem doing summation

I'm having troubles doing this sum as part of a past paper im doing for exam friday.

Theres an image of a region of the graph y = 3^x between 0 and 1 with a set of strips shown beneath the curve with width 'h' so that nh = 1 where n is the number of strips

i) By using the set of rectangles indicated on the diagram show that A > 2h/(3^h - 1) where A is area of the curve between 0 and 1

Now, summing the areas of the strips, i get h(1+3^h + 3^2h .... + 3^[h(n-1)]) but im having trouble trying to get that to be 2h/(3^h - 1)

I know how to sum a geometric series from way back in the modules, but if i plug it in here i get.

Have i done something wrong, because i cannot find a way of equating that to the expression given.

Your sum is:

But nh=1 => h=1/n

I don't get how you did it

Last edited by Identity (2008-05-19 04:38:42)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: problem doing summation

i see now, i was trying to put it into a form where the ratio was a single number, i had r = 3^h at first, but thought, that can't be right, forgetting that 'h' is known thanks.

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The Beginning Of All Things To End.
The End Of All Things To Come.