Question

prove that when 1 is added to the sum of n terms of the series {8+16+24+...} , the result will be a perfect square.​

Answer 1

Answer:

try putting values in it you will get a perfect square

used sn as formula

mark me as brainliest

Answer 2

Answer:

Sn + 1 = 4n^2 + 4n + 1

Step-by-step explanation:

t1 = 8 c.d. = 16 - 8 = 8

Therefore ,

Sn = n/2 [ 2 t1 + (n - 1) d]

=> Sn = n/2 [ 16 + (n - 1) 8]

=> Sn = 4n (2 + n - 1)

=> Sn = 4n^2 + 4n

Now , A.T.P ,

=> Sn + 1 = 4n^2 + 4n + 1

=> Sn + 1 = (2n)^2 + 2 . 2n . 1 + (1)^2

Therefore ,

Sn + 1 = (2n + 1)^2

As (2n + 1) is always even for all (+)ve integer values of ' n ' .

Therefore , (2n + 1)^2 is a perfect square.

HOPE IT WILL HELP YOU!

THANK YOU.