Question

if Tn= 2n-1 than S2
​

Answer 1

Answer:

Tn = 2n - 1

let n be 1

T1 = 2 - 1 = 1

Let n be 2

T2 = 4 - 1 = 3

Common difference ( d ) Will be 2

sn = \frac{n}{2} (2a + (n - 1)d)sn=

2

n

(2a+(n−1)d)

sn = \frac{n}{2} (2 + (n - 1) \times 2)sn=

2

n

(2+(n−1)×2)

sn = \frac{n}{2} (2 + 2n - 2)sn=

2

n

(2+2n−2)

sn = \frac{n}{2} (2n)sn=

2

n

(2n)

sn = \frac{2 {n}^{2} }{2}sn=

2

2n

2

Sn = n²

Answer 2

Answer:

S2=4 is the correct answer