Genius come at me.

[[MPGH]Wyo]

Can you explain me how to do this please?

a + (a+d)+(a+2d)+...+[a+(n-1)d] = (n/2)[2a+(n-1)d]

[[MPGH]Austin]

uhm..

carry the 2, minus the ....


OK. THE ANSWER IS 42.

[[MPGH]Wyo]

Real answers Please @Austin

[Alen]

[4chan]

Alen I did this way.








k be ANY value of n
Pk is then proposition that :-
a + ( a + d ) + --------[ a + ( k - 1) d ] = (k/2) [ 2a + (k - 1) d ]

Have now to show that this is true for k = 1 and k = k + 1

Consider k = 1
---------------------
LHS = a
RHS = (1/2) (2a) = a
Thus true for k = 1

Consider k = k + 1
----------------------------
a + ( a + d ) + --------[ a + ( k) d ] = (k+1)/2) [ 2a + (k) d ]
Have now to show this to be true:-

Now
a + ( a + d ) + ---[a + (k - 1)d] = (k/2) [2a + (k - 1) d ]___IS true
a+(a + d)+ ---[a + (k)d] = (k/2) [ 2a + (k - 1) d ] + [a + kd]

Consider RHS :-
(k/2) [ 2a + (k - 1) d ] + [a + kd]
ka + k(k-1)d/2 + a + kd
(k + 1)a + kd [ (k - 1)/2 + 1 ]
(k + 1)a + kd [ (k + 1)/2 ]
[ (k + 1)/2 ] [ 2a + kd ]
Thus P (k + 1) is true

We now have:-
Pk true
P1 true
P k + 1 true
Thus true for all k
Thus Pn is true
@Alen

[Alen]

That's called mathematical induction and yeah, that's how one would go to prove such stuff.

[4chan]

Yeah, fuck University they send alot of homeworks.

[Alen]

Yeah, fuck University they send alot of homeworks.

I sincerely doubt that's university grade, mathematical induction is usually somewhere around grade 10-12 or something?

[4chan]

Not in Spain I wanna be an architect