Question

Find the sum.

Answer 1

\[1 + 3 + 5 +...+ (2n -1)\]

Answer 2

Use\[\Large
\sum_{k=1}^n k= \frac{n(n+1)}{2}
\]

Answer 3

i know how to do these types of problems, i just got stuck on this one.

Answer 4

@Directrix no its not like how you mention.

Answer 5

\[
\Large
\sum_{k=1}^n [2k-1]=\sum_{k=1}^n 2k-\sum_{k=1}^n1
\]

Answer 6

it's an AP

Answer 7

@mathslover what?

Answer 8

Then, you want the sum in terms of n, I suppose.

Answer 9

why don't you use the AP sum formula ?

See , AP is arithmetic progression

Answer 10

Here we have d (common difference) as 2 ..

Hey, before I get back to the complete work started, can you tell me that have you learnt Arithmetic Progression @some_someone

Answer 11

@mathslover yes i have.

Answer 12

fine, so can you tell me the formula for Sum of an AP

Answer 13

i am not sure, can you tell it to me, and i will tell you if i have used it.

Answer 14

OK! it is :
\[\large{S_n = \frac{n}{2} (2a + (n-1)d)}\]

where n is the number of observations .
a is the first term
d is the common difference

Answer 15

oh its that one, yes i know it ^.^

Answer 16

hmn. Well @some_someone lemme know exactly that which topic were you doing when you encountered with this question ??

Answer 17

ok however, i use also
\[S _{n} = \frac{ n }{ 2 }(a _{1} + a _{n})\]

Answer 18

that's also right..

Answer 19

the sum of the first n odd number is \(n^2\)

Answer 20

yeah i know, that is the one i have been using

Answer 21

OK I got it.

Answer 22

let the number of observations be "k"

Answer 23

you can see it with a picture

Answer 24

@satellite73 , the sum is \[n^2 \] and that is the answer i just dont know how to get to that answer?

Answer 25

then we have :
\[\large{S_k = \frac{k}{2} (2(1) + (k-1)2)}\]
\[\large{S_k = \frac{k}{2} (2(1+k-1))}\]
\[\large{S_k = k (1 + k -1 )}\]
\[\large{S_k = k ( \cancel{1} + k - \cancel{1})}\]
\[\large{S_k = k^2}\]
Proved

Answer 26

pattern is clear for one thing

\[1=1\]\[1+3=4\]\[1+3+5=9\]\[1+3+5+7=16\]

Answer 27

@satellite73
instead of mathematical induction, we can also prove it algebraically

Answer 28

you don't even need algebra, a picture will do it

Answer 29

Yes but I believe in proving it algebraically .

well, @some_someone did you get it ?

Answer 30

can u use the formula i gave you to show me?

Answer 31

Yes @some_someone I used that

Answer 32

See I have :
\[\large{S_n = \frac{n}{2} (2a + (n-1)d)}\]

Right?

Answer 33

no

Answer 34

why ?

Answer 35

\[\large{S_n = \frac{n}{2} (a_1 + a_n)}\] ? this one ?

Answer 36

yes

Answer 37

OK now , put a_1 = 1 and a_nn = 2n-1 , note that we are given with this information .

Can you solve it ?

Answer 38

hmmm... idk

Answer 39

how about this
\[n^2=\sum_{k=1}^nk^2-(k-1)^2=\sum_1^n k^2-(k^2-2k+1)=\sum_{k=1}^n 2k-1\]

Answer 40

i dont get it?

Answer 41

the line i wrote or above it?

Answer 42

urs?

Answer 43

OK satellite sir will explain you, I think he is good at that. I will recommend you @some_someone to understand the way @satellite73 sir is explaining.

Answer 44

\[\sum_{k=1}^nk^2-(k-1)^2\] is a telescoping sum
the only term that survives is the last one , \(n^2\)

Answer 45

the rest is algebra, showing that
\[k^2-(k-1)^2=2k-1\]

Answer 46

@satellite73
@mathslover
so i did the following
\[S _{n} = \frac{ n }{ 2 }(a _{1} + a _{n})\]
\[S _{n} = \frac{ n }{ 2 }(1 + 2n-1)\]
\[S _{n} = \frac{ n }{ 2 }(2n)\]
am i doing it right?

Answer 47

looks good to me

Answer 48

OK you're right @some_someone but did you get the method suggested by satellite sir?

Answer 49

no :(

Answer 50

Hmn... leave that ,,, you are OK in AP ?

Answer 51

yep :)
thanx everyone :D