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In the prior lecture we have shown how Gauss solved the problem of finding the sum 1+2+3+
...up to 100 in Gauss solution we reduced the problem of finding the sum of different
natural numbers to the problem of finding the sum of 50 equal numbers. It is very natural
in mathematics to generalized concepts and results a natural small generalization of
the prior result will be to find the sum 1+2+3 + up to n of the first n natural numbers.
It is also a natural impulse to try to solve similar problems with similar solutions. In
this case we will try to solve the problem using the same idea as Gauss, with a very
small modification.
As in the prior lecture let us call T sub n equal 1 +2 + 3 + up to n the n triangular
number. we can easily rearrange the order of the elements in the addition in reverse
order. Therefore we can write. T sub n equal n+ n-1 + n-2 + plus 1
if we add these two equalities we get 2 times T sub n equal n=1 plus n+1 plus n+1
all of them are n+1 and as in Gauss solution to the problem we have found again that adding
a number from the beginning of the sequence to one from the end the sum stays constant.
The right hand side is also very easy to compute. We have n of those terms. Therefore T sub
n equal n times n+1 over 2. This solution was easy to find because the solution is base
on the same idea as Gauss's solution or the special case
1+2+3+ ... up to 100. We already pointed out why Gauss's solution works and the same is
true here. The solution works by translating the problem of finding the sum of different
numbers to finding the sum of equal numbers and we are able to produce the equal numbers
by conveniently rearranging the numbers in the sequence. In both cases we are dealing
with sequences of consecutive natural numbers. Can we generalized this a bit more? Yes we
can. What if instead of the sequence that starts
on 1 we get a sequence that starts on a number a sub 1 and we obtain the next element by
adding a constant natural number d so the first element of this progression will be
a sub 1 the second element is a sub 2 equal to a sub 1 plus d and a sub 3 equal a sub
1 plus 2 times d and so on until a sub n equal a sub 1 plus n -1 times d. This progression
is a bit more general than the sequence of natural numbers. First it does start on an
arbitrary number and the difference between two consecutive terms is d instead of 1. Progressions
that satisfy this conditions are called arithmetic progressions.
Examples of arithmetic progressions are
1,2,3, up to 100 our special case. In this case the first element is 1 and the value
for d is 1 and another example is 2,4,6 ... up to 200 in this other example
the first element is 2 and the increment is 2 so we obtain each term by adding 2.
Can we find a formula for the sum of the general arithmetic progression? That is to find
a sub 1 plus a sub 2 plus a sub 3 plus up to a sub n where a sub k is equal to a sub
1 plus k-1 times d for k from 1 to n. We find this formula we may be able to encounter
some invariant as before by adding terms from the beginning and the end of the progression.
We should be able to prove that a sub 1 plus a sub n is equal to a sub 2 plus a sub n-1
and the results is constant. That is we should prove that
a sub k + a sub n - (k-1) equal a sub 1 plus a sub n for k from one to n. We know by our
definition that a sub k equals a sub 1 plus d times k minus 1 and also a sub n - k-1 equals
a sub 1 plus d times n - (k-1) -1 adding both equalities we get a sub k plus a sub n - (k-1)
is equal to a sub 1 plus d times k-1 plus a sub 1 + d times n-(k-1)-1) and this is equal
to a sub 1 plus a sub 1 plus d times k-1 + n-k+ 1-1 and this is equal to a sub 1 plus
a sub n therefore if we call S sub n equal a sub 1 plus a sub 2 plus a sub 3 plus up
to a sub n and rearranging the terms as before and adding both equalities
we get two times S sub n equal a sub 1 + a sub n + a sub 2 plus a sub n-1 plus a sub
n plus a sub 1. But applying the prior result we get that two times S sub n equals a sub
1 plus a sub n and so on until the last one since there are n of those equal terms we
get the S sub n equal a sub 1 plus a sub n times n over 2. Finally we will propose some
problems where you can apply similar solutions. Please send the solutions to solutions@isallaboutmath.com