Question 795588
1- the sum of the positive four digit odd integers:

Let {{{x}}} be and {{{odd}}} number. Then {{{x}}}, {{{x+2}}},{{{x+4}}},{{{x+6}}}   are four consecutive odd numbers, and their sum is

{{{x+(x+2)+(x+4)+(x+6)=4x+12}}}

2- write in sigma notation {{{12+3+3/4+3/16+3/64}}}

the general formula for this sequence is {{{a[n ]= 3*4^(2-n)}}} (for n=1,2,....)

you have five terms, so {{{n}}} starts with one and ends with five

{{{sum(3*4^(2-n),n=1,n=5)=12+3+3/4+3/16+3/64=15+(63/64)}}}