Question

(1^2+1) +(2^2+2) +(3^2+3) +........... +(n^2+n) =?
1) n(n+1) (n+2) /6
2) n(n+1) (n+2) /3
3) n(n+1) (2n+1) /3
4) n(n+1) (2n+1)/3+(n+1) /2​

Answer 1

Answer:

ANSWER

Let n=1

Then 

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

On the other hand,

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

So these expresions 

coincide.

Supose that we have proved the 

identity for all k

so 

in particular, for k=n we have that 

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

We have to prove the identity for 

k=n+1, that 

is

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

Notice that by induction

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

= n/2[2a+(n-1)d] + [a+nd] 

=na + n(n-1) d/2 + a + nd 

=(n+1) a + (n^2/2 - n/2 + n) d 

=(n+1) a + (n^2/2 + n/2) d 

=(n+1) a + (n+1) n d /2

=(n+1)/2 [ 2 a + n d] 

Hence prove !

Answer 2

♡This helps u dear♡

nice day