Prove: \[\huge S_n = \frac{n(a_n + a_1)}2\]

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Prove: \[\huge S_n = \frac{n(a_n + a_1)}2\]

Mathematics
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it's an easy proof....but it's still nice to see some professional do it
i wrote that formula wrong...
See \[\S_n = \cfrac{n}{2} (2a + (n-1)d)\]

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Other answers:

and..?
Now put 2a +(n-1)d = a + a+(n-1)d = a_1 + a_n
You get ur answer !
i don't think that proves anything...that was formula transformation
see a_n = a+(n-1)d
That really does //////////// :)
You don't want to use the first formula ?
can you prove a_n = a + (n-1)d?
I think yes
yep
i think you;re confused....this is proving...not substitution
suppose we have 1,2,3,4.....100 adding all, we can see it like this (1+100) + (2+99) .... ( 50 + 51) => (101)*50 = (1 + 100) * (100/2) you may generalize this..
And u asked for a proof ;)
that wasn't a proof @vishweshshrimali5 ....it was formula transformation
and substitution isn't proving either.... @shubhamsrg
Whatever........... leave it I can prove that without using a_n formula
substitution ? o.O
But you will have to accept that a_n = a+(n-1)d That is basic def. of AP
I am also surprised @shubhamsrg O.o
i am not! ;)
;)
are any of you familiar with mathematical induction?
Yes
You want that proof........
what both of you did was substitution (which is NEVER accepted as proofs)
Let P(n) be the given statement
no. i want other proof
induction is too dull and boring and predictable
you're amazing you know that ?
not you ! ;)
@lgbasallote please give us a "hint" of the proof u want
actually...it's surprising @vishweshshrimali5 does know induction...
hint? lol..
any proof as long as it's valid and not induction
induction is one of the many great tools we have in mathematics today!
preferrably, i like to see direct proof and contradiction
i like deduction much more
Ok lets try u want to deduce a formula from a continuous pattern
time for the site to go down....
bye
bye
so i suppose i'll just ask this again in the fture
will check out later
yep

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