let s sub k for k is equal to one all the eay to one hundred denote the sum for the infinite geometric series
whose first term is k minus one over k factorial and the common ratio is 1 over k. Then the value of
100 squared over 100 factorial plus the sum from k equals one to one hundred of this thing over here
the absolute value of this thing over here k2 minus 3k + 1 times s sub k is. So this looks like a pretty
daunting thing but I guess a good place to start would just to be find a simple way to express s sub
k. So if we were to write s sub k. s sub k is equal to the first term is k-1 over k factorial and then t
he common ratio is one over k. So it's going to be one over k to the 0 plus one over k to the 1 plus
1 over k squared. It is just going to keep going on forever, it is an infinite geometric series. Now t
there is a very well known formula for the sum of an infinite geometric series. I enjoy the proof of i
t so much that we're going to do it again. and I've done it multiple times already but its such a simple
proof that it is good to do again, if you're under time pressure then you could just use the formula.
but if we say, lets just say some other sum, lets call this thing just s. s is equal to 1/k^0 + 1/k^1
+ 1/k^2 all the way to infinite. Now lets multiply 1/k * s, which is going to be 1/k times all of these t
terms. So I multiply 1/k times this term I'm going to get (1/k)^1. When I multiply 1/k times this term
I'm going to get 1/k to the second power all of the way on to infinite. So if I were to subtract this f
rom that so it's coefficient is one 1 - 1/k * s would be equal to all of this stuff minus all of this
stuff. That is going to cancel with that and that is going to cancel with that and we're just going t
to have the 0 term here which is to the first power. 1/k ^ 0 is going to be equal to 1, or this whole s
sum is going to be equal to 1 over / 1-1/k. So we can re-write s sub k s sub k is equal to k - 1 over
k! times before I write it up here lets see if we can simplify this. If I want to simplify this I can
multiple the numerator and the denominator by k. If do this I am going to have k in the denominator a
nd the denominator becomes k -1. So this right here is k over k - 1 and these will cancel out so this
we have to be careful when we cancel out as this kind of precluded, while I will talk about that in a

級數和的例子 (英)

級數和的例子 (英) : 2010 IIT JEE Paper 1 Problem 54 Series Sum