Question 785568
I would add those 9 numbers twice and then divide by 2.
{{{((1+2+3+4+5+6+7+8+9)+(1+2+3+4+5+6+7+8+9))/2}}}
Rearranging them in pairs, matching them head to tail, I get nine sums added up
= {{{((1+9)+(2+8)+(3+7)+(4+6)+(5+5)+(6+4)+(7+3)+(8+2)+(9+1))/2}}}
and each of them adds to 10
= {{{(10+10+10+10+10+10+10+10+10)/2=9*10/2=9*5=highlight(45)}}}
 
Your teacher will say that the sequence of numbers from 1 to 9 is an arithmetic sequence (or an arityhmetic progression, depending on where you go to school), and to apply formulas like
{{{a[n+1]=a[n]+d}}},
{{{a[n]=a[1]+(n-1)*d}}}
{{{S[n]=sum(a[i],i=1,i=n)=n(a[1]+a[n])/2}}} and
{{{S[n]=sum(a[i],i=1,i=n)=(2a[1]+(n-1)d)n/2}}}
and that for the numbers from 1 to 9,
{{{a[1]=1, {{{d=1}}}, {{{n=9}}}, {{{a[9]=9}}}
If I had to deal with that, I would use
{{{S[n]=sum(a[i],i=1,i=n)=n(a[1]+a[n])/2}}}
and calculate
{{{S[9]=sum(a[i],i=1,i=9)=9(1+9)/2=9*10/2=highlight(45)}}}
which is exactly what I did above (but without the formulas).