Problem 25, section 3.2Ugh. This problem. Well, lets see. We want to find a sum formsummationdisplayk=0floorleft√kfloorrightFor simplicity, we’ll letn=floorleft√mfloorright -1, so thatfloorleft√mfloorright=n+ 1. Why isnsignificant? Well, if you right out the summsummationdisplayk=0floorleft√kfloorright= 0 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 3 + 3 +. . .you’ll notice thatnis the second to last term that appears (before we startaddingfloorleft√mfloorright. This is important because we don’t know how many times weaddfloorleft√mfloorright. But we do now how many times we add everything below it. As youcan see above, this fromula looks a lot like∑mk=0(2k+ 1)k. But is isn’t quitethe same, because we may not addfloorleft√mfloorrightas much as possible. Thus, we needto break this sum into two parts:nsummationdisplayi=0(2k+ 1)k+summationdisplayfloorleft√mfloorrightYou’ll notice the sum on the left is relatively straight forward, and if youwork it out, it is equal ton(n+1)(2n+1)6+n(n+1)2. Thus we only need to figure
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