rosen(sets relation function)

Question

Find a formula for $\sum_{k=0}^{m}G.I.F(\sqrt{k})$, when m is a positive integer (where G.I.F is greatest integer function or floor function).

Comments

for 1 to 3,GIF will be 1,for 4 to 8 it will be 2,for 9 to 15,it will be 3,for 16 to 24.it will be 4....and so on

nhence,summation will be like -

3*1 + 5*2 + 7*3 + 9*4 +...[m-(n+1)² +1] *(n+1) where n is the second term of every term

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@Akriti, Mam how u've got dis ans??. Please can u explain in simple steps. :)

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i dun know how to solve this equation further as it two different AP.
and i got this simply,sqrt(1)=1
sqrt(2) =1 .41 and hence floor will be 1
sqrt(3) =1.73 and hence floor will be 1
sqrt(4) = 2 and hence floor will be 2
sqrt(5) =2.23 and hence floor will be 2
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sqrt(8)=2.82 and hence floor should be 2
similarly solve like this

Answer 1

If we write down the first few terms of this sum we notice a pattern.

For 1 to 3 GIF will be 1 ,

For 4 to 8 GIF will be 2 ,

For 9 to 15 GIF will be 3 ,

For 16 to 24 GIF will be 4

It starts (1 +1+1) + (2 + 2 + 2 +2 + 2) + (3 + 3 + 3 + 3 + 3 + 3 + 3) + · · ·. There are three l's, then five 2's, then seven 3's, and so on;
In general there are $(i+1)^{2} - i^{2}$ = 2i + 1 copies of i. So we need to sum i(2i + 1) for an appropriate range of values for i.

We must find this range. It gets a little messy at the end if m is such that the sequence stops before a complete range of the last value is present. Let n = floor(√m) - 1. Then there are n + 1 blocks, and $(n+1)^{2}$ - 1 is where the next-to-last block ends.

The sum of those complete blocks is
$\sum_{i=1}^{n}$ i(2i + 1) = $\sum_{i=1}^{n}$ 2$i^{2}$ + i = n(n + 1)(2n + 1)/3 + n(n + 1)/2.

The remaining terms in our summation all have the value n + 1 and the number of them present is m - ($(n + 1)^{2}$ - 1).

Our final answer is therefore
n(n + 1)(2n + 1)/3 + n(n + 1)/2 + (n + 1)(m - $(n + 1)^{2}$ + 1).