Question

If tn = 3n+1, then Sn =



​

Answer 1

tn=3n+1

Sn-Sn-1=tn

Sn-Sn-1=3n+1

take out the first term

T1=4

second term

T2=7

D=7-4=3

SN=n/2(2×4+(n-1)3)

Sn=n/2(5+3n)

Sn=5n/2+3n^2/2

Sn=(3n^2+5n)/2

Answer 2

If $t_{n}=3n+1$, then $S_{n}=\dfrac{n}{2}(3n+5)$.

Step-by-step explanation:

Here, $t_{n}=3n+1$

Then,

• $t_{1}=3(1)+1=3+1=4$

• $t_{2}=3(2)+1=6+1=7$

• $t_{3}=3(3)+1=9+1=10$

Clearly, $t_{2}-t_{1}=t_{3}-t_{2}=3$ and thus the terms are in Arithmetic Progression, whose first term $(t_{1})$ is $4$ and common difference $(d)$ is $3$.

Now, the sum of first $n$ terms is

$S_{n}=\dfrac{n}{2}[2t_{1}+(n-1)d]$

$=\dfrac{n}{2}[2(4)+(n-1)(3)$

$=\dfrac{n}{2}[8+3n-3]$

$=\dfrac{n}{2}(3n+5)$

Note:

You could also use the following formula to find the required sum:

$\quad S_{n}=\dfrac{n}{2}(t_{1}+t_{n})$