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I am solving this equation:

Image of the equation
My issue arises on line 2 where we have (n + 1 - 2 + 1)(n + 1 + 2)/2. This is what i understand. We have the formula n! = n(n+1)/2. Subbing values into the equation yields us with 10[(n+1)(n+1 + 1)/2] however we need to account for the fact we are starting at j = 2. I would then do this by 10[(n+1-2)(n+1+1)/2]. What I dont understand is why in the solution they have given

  • (n+1 - 2 - 1 +1) the -1 + 1
  • (n + 1 + 2) the + 2

How did they make that logical jump and what were the steps/thought process in doing so.

JMP
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archhmod
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2 Answers2

4

in line 2 the $\frac{(n+1+2)}{2}$ part is the average of the terms being summed. In this problem this is calculated by adding the first and last term and dividing by 2. The $(n+1-2+1)$ part is the number of terms being summed. This is calculated by subtracting the index of the first term (2) from the index of the last term (n+1) and adding one.

The sum is equal to the average value of each term multiplied by the number of terms.

Also, $n!=\frac{n(n+1)}{2}$ is not true. The correct equation is $\sum_{i=1}^n i = \frac{n(n+1)}{2}$

user781829
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  • I see, the first part of the equation makes sense. n+1 is almost like how many times the formula is to be executed. Our starting point is 2 so to compensate we subtract 2 and add 1. Where I am still lost is how we have (n+1+2). In my head this should be our number of iterations n+1 and then +1, why is it +2? – archhmod May 01 '20 at 06:18
  • If we started at 1 the average term would be (n+1+2)/2, but since we start at 2 the average term will be a bit larger. That is the intuition. In general, when adding a bunch of successive integers the average term will always be the first term plus the last term divided by 2. – user781829 May 01 '20 at 06:27
  • Yeah I had a brain fart dude, makes sense, the average term must be shifted across by 2 because we are starting at 2. Lets say j = 0, how would this change our answer? – archhmod May 01 '20 at 06:28
  • the average term in the sum $\sum_{i=0}^{n+1} i$ would be $\frac{n+1}{2}$. The first term (0) plus the last term (n+1) all divided by 2. (in my previous comment it is unclear, but the addition has to be done before dividing by 2) – user781829 May 01 '20 at 06:43
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$n+1-2+1$ gives the number of terms, which is the number of integers in the inclusive range $[2,n+1]$.

$\dfrac{n+1+2}{2}$ gives the average term, which is the largest term ($n+1$) plus the smallest term ($2$) divided by $2$.

JMP
  • 21,771
  • I see, can you explain to me where in the second part of the equation where how does (n+1+2) come to be? Why isnt it simply (n+1+1)? – archhmod May 01 '20 at 06:19