Question

use Gauss's approach to find the following sums (do not use formulas).
1+3+5+7+...+997
what is the sum of the sequence?

Answer 1

thisis realy confusing me its like i almost keep understanding it then i dont understand and get confused!

Answer 2

gauss a approach is to add up forwards and backwards
\[S=1+3+5+...+997\]
\[S=997 + 995 + ... + 3 + 1\]
\[2S=998 998 + ... + 998= 499\times 998\]
\[ S=499^2\]

Answer 3

ok that was a typo, second line should have been
\[S= 998 + 998 + ... + 998=499\times 998\] but this is a silly way to do this

Answer 4

1+3+5++.....+997
997+995+993+....+1
=998+998+......+998(499 times)
so =998*499/2

Answer 5

it is always true that
\[\sum_{k=1}^n(2k-1)=n^2\]

Answer 6

where did the 499 come from??

Answer 7

good question is how do we know 499

Answer 8

oh right, you just asked it !

Answer 9

all the numbers are odd, so they all look like
\[2k-1\] for k = 1,2,3,... and you want to know how many terms there are, because that is what you have to multiply by. set
\[2k-1=997\]
\[2k=998\]
\[k=998\div 2=499\]

Answer 10

omg thats so confusing!!

Answer 11

however, it should be entirely clear that the sum the first n odd numbers is n^2

\[1=1^2\]
\[1+3=2^2\]
\[1+3+5=3^2\]
\[1+3+5+7=4^2\]
\[1+3+5+7=5^2\] etc

Answer 12

can i give u another one and show me again with more detail steps

Answer 13

sure

Answer 14

48+49+50+51+... +133

Answer 15

but basically it comes down to figuring out how many terms you have. and although summations can be confusing i am sure you are not confused by how to solve
\[2k-1=997\] for k

Answer 16

again using gauss?

Answer 17

is their an easier way to solve these without so many steps?

Answer 18

yes same process

Answer 19

it is as the last example. one half the number of terms times (first plus last term)

Answer 20

i really need to understnad this by today cuz this assignment was due yesterday and its so long and i still didnt finish it i have to submit it today!

Answer 21

in the last example the first term is 48, last term is 133 and their sum is
\[48+133=181\]
now how many terms are there?
should get 86 terms. so your answer will be

\[\frac{1}{2}\times 86\times 181\]

Answer 22

so 86x181/2 ?

Answer 23

yup

Answer 24

thank you!

Answer 25

its like it makes sense but at the same time i dont get it lol

Answer 26

so what do you need all together to solve these additions :
first term
\[a_1\]
last term
\[a_n\]
number of terms.
\[n\] then it will be
\[S =\frac{n}{2}(a_1+a_n)\]

Answer 27

you will get it if you do a bunch of examples