If S_n = n² + 2n, find T_n

From the question we know that, S_n = n² + 2n ……. (1)

We know that, S_n = × (2a + (n – 1)d)

= na +

= na + (n² – n) ×

S_n = na + n² × – n × ……. (2)

As we know that, (1) and (2) are both sum of the same arithmetic progression

So we can equate them to each other.

So, we have,

⇒ n² + 2n = na + n² × – n ×

⇒ n² + 2n = n² × + na – n ×

Now we will equate the coefficients of “n²” and “n” on both the sides of the equal to sign.

So, we get,

For coefficients of n² :

⇒ 1 =

⇒ d=2

Now, For coefficients of n:

⇒ 2 = a –

From above we have that d=2,

So we can say that,

⇒ 2 = a –

⇒ 2 = a – 1

⇒ a=3

So, now, we know that, T_n = a + (n – 1)d

Now we put values of a and n in the above equation,

We get,

⇒ T_n = 3 + (n – 1)2

⇒ T_n = 3 + 2n – 2

⇒ T_n = 2n + 1

∴ for the given value of S_n = n² + 2n, we have T_n = 2n + 1.