Question

Find the sum of 2n terms of the series 1^2-2^2+3^2-4^2+5^2-6^2...

Answer 1

We have series : 12 - 22 + 32 - 42 + 52 - 62 + . . .

We can write our series , As :

⇒( 12 - 22 ) + ( 32 - 42 ) + ( 52 - 62 ) + . . .

⇒[ ( 1 - 2 ) ( 1 + 2 ) ] + [ ( 3 - 4 ) ( 3 + 4 ) ] + [ ( 5 - 6 ) ( 5 + 6 ) ]+ . . .

⇒( -1 ) ( 3 ) + ( -1 ) ( 7 ) + ( -1 ) ( 11 )+ . . .

⇒( - 3 ) + ( - 7 ) + ( - 11 ) + . . .

Here first term = a = - 3

And

Common difference = d = - 7 - ( - 3 ) = - 7 + 3 = - 4 , - 11 - ( - 7 ) = -11 + 7 = - 4

We have formula for sum of n terms in A.P. , As :

Sn = n2[2a + (n − 1 )d]
Here
n = Number of terms = 2 n
a = first term = - 3
and
d = Common difference = - 4

So,

S2n = 2n2[2( −3) + (2n − 1 )( − 4 )] = n[ − 6 + ( −8n + 4 )]= n[ − 6 + −8n + 4 ]= n[ −8n − 2 ] = − 8n2 − 2n = −( 8n2 + 2n ) ( Ans )
Hope this information will clear your doubts about Arithmetic Progressions.