Question 868013
The first 100,000 odd numbers are the odd numbers from
{{{2*1-1=2-1=1}}} to {{{2*100}}}{{{",000"-1=200}}}{{{",000"-1=199}}}{{{",999"}}} :
1, 3, 5, .... 199,993, 199,995, 199,997, 199,999.
The sum of such an arithmetic sequence
(or arithmetic progression, if that's what it's called in your class)
is 100,000 =100000={{{10^5}}} (the number of terms in that sum)
times half the sum of the first and last terms,
{{{(1+199999)/2=200000/2=100000=10^5}}}= 100,000.
So that sum is {{{100000*100000}}}= 10,000,000,000,
or {{{10^5*10^5=10^10}}} .
Your teacher may want to see formulas.
I do not know what formulas you were taught,
and I believe in understanding concepts rather than memorizing and blindly applying formulas,
but I can throw a few formulas for you to use as needed.
The same concepts can be expressed in many different ways.
An arithmetic sequence has a first term {{{a[1]}}},
and each term is the term before plus a common difference {{{d}}} ,
which can be written as
{{{a[n]=a[n-1]+d}}} .
As a consequence
{{{a[2]=a[1]+d}}}
{{{a[3]=a[1]+2d}}}
{{{a[n]=a[1]+(n-1)d}}} ,
and of course, {{{a[n-1]=a[1]+(n-2)d}}}
The sum of the first {{{n}}} terms is
{{{S[n]=a[1]+a[2]+"..."+a[n-1]+a[n]}}}={{{(1/2)((a[1]+a[2]+"..."+a[n-1]+a[n])+(a[n]+a[n-1]+"..."+a[2]+a[1]))=(1/2)((a[1]+a[n])+(a[2]+a[n-1])+"..."+(a[n-1]+a[2])+(a[n]+a[1]))}}}
That last expression contains a sum of {{{n}}} 2-term sums, and all the 2-term sums are equal:
{{{a[1]+a[n]=a[1]+(a[1]+(n-1)d)=a[1]+a[1]+(n-1)d=2a[1]+(n-1)d}}} ,
{{{a[2]+a[n-1]=(a[1]+d)+(a[1]+(n-2)d)=a[1]+d+a[1]+(n-2)d=2a[1]+(n-1)d}}} .
So {{{S[n]=(1/2)(a[1]+a[n])*n=n*((a[1]+a[n])/2)}}} .
That is the number of terms times half the sum of the first and last terms,
or (in other words) {{{n}}} times the average of the first and last terms.
An alternate formula is
{{{S[n]=(1/2)(2a[1]+(n-1)d)*n}}} .
 
If {{{x}}} is one of two natural numbers whose sum is {{{41}}} ,
then the other number is {{{41-x}}} ,
and their product {{{y}}} is
{{{y=(41-x)x=41x-x^2}}}.
Traditionally, we prefer to rearrange the terms in the order {{{y=-x^2+41x}}} .
We notice that {{{y}}} is a quadratic function of {{{x}}} ,
and its graph would look like this,
{{{graph(300,300,-5,45,-500,500,-x^2+41x)}}} , with a maximum.
To find the maximum we can "complete the square":
{{{y=-x^2+41x}}}<--->{{{y=-(x^2-41x+(41/2)^2)+(41/2)^2}}}<--->{{{y=-(x-41/2)^2+(41/2)^2}}}
Alternately, we can use the formula that says that the maximum for {{{y=ax^2+bx+c}}} occurs at {{{x=-b/2a}}} .
If we could use any rational value for {{{x}}} ,
the maximum would occur for {{{x=41/2=20.5}}} ,
where we would have {{{y=(41/2)^2=41^2/2^2=1681/4=420.25}}} .
However, since {{{x}}} must be a natural number,
the maximum value for the product will happen for the natural numbers {{{x}}} closest to {{{20.5}}} :
{{{x=20}}} ---> {{{41-x=41-20=21}}}
and {{{x=21}}} ---> {{{41-x=41-21=20}}} .
IN either case, the product will be
{{{20*21=highlight(420)}}}