Question

prove that the sum of any number of the arithmetic sequence 16,24,32 ect starting from the first add to 9 gives a perfect square ​

Answer 1

Given,

A.P = 16,24,32,...

Sum of n terms can be determined by the equation,

$\boxed{\bold{S_n = \frac{n}{2}[2a+(n-1)d]}}$

Here,

• a = 16
• d = 8

$\implies\bold{S_n = \frac{n}{2}[(2\times16)+(n-1)8]}$

$\implies\bold{S_n = \frac{n}{2}[32+8n-8]}$

$\implies\bold{S_n = \frac{n}{2}[8n+24]}$

$\implies\bold{S_n = n(4n+12)}$

$\implies\bold{S_n = 4n^2+12n}$

Adding 9 to both the sides,

$\implies\bold{S_n+9 = 4n^2+12n+9}$

$\implies\bold{S_n+9 = (2n)^2+(2\times2n\times3)+(3)^2}$

$\implies\bold{S_n+9 = (2n+3)^2}$

Hence, $\bold{S_n+9}$  is a perfect square

Hence Proved........