Question

find the sum of n terms of the series whose nth term is n^2 + 3n + 1​

Answer 1

According to the question:

$\sf a_{n} = n ^{2} + 3n + 1$

Thus

$\sf⟹ a_{n} = a + (n - 1)d = n²+3n+1$

$\sf ⟹a_{1} =a=(1)²+3(1)+1$

$\sf ⟹a=1+3+1$

$\sf a= 5$

Thus 1st term is 5

$\sf ⟹a_{1} =a=(2)²+3(2)+1$

$\sf ⟹a_{2}=4+6+1$

$\sf a_{2}= 11$

Thus 2nd term is 11

Therefore Common Difference

$\sf d=a_{2}-a_{1}$

$\sf d=11-5$

$\sf d=6$

Now

$\sf S_{n}= \frac{n}{2} (2a+(n-1)d$

$\sf S_{n}=\frac{n}{2} (2(5)+(n-1)6$

$\sf⟹\frac{n}{2} (10+6n-6)$

$\sf ⟹S_{n} = \frac{n}{2} (6n+4)$

$\sf ⟹S_{n}=\frac{6n²+4n}{2}$

⟹ Sn = 4n²+2n

Answer 2

Answer:

sn

Step-by-step explanation:

According to the question:

\sf a_{n} = n ^{2} + 3n + 1a

n

=n

2

+3n+1

Thus

\sf⟹ a_{n} = a + (n - 1)d = n²+3n+1⟹a

n

=a+(n−1)d=n²+3n+1

\sf ⟹a_{1} =a=(1)²+3(1)+1⟹a

1

=a=(1)²+3(1)+1

\sf ⟹a=1+3+1⟹a=1+3+1

\sf a= 5a=5

Thus 1st term is 5

\sf ⟹a_{1} =a=(2)²+3(2)+1⟹a

1

=a=(2)²+3(2)+1

\sf ⟹a_{2}=4+6+1⟹a

2

=4+6+1

\sf a_{2}= 11a

2

=11

Thus 2nd term is 11

Therefore Common Difference

\sf d=a_{2}-a_{1}d=a

2

−a

1

\sf d=11-5d=11−5

\sf d=6d=6

Now

\sf S_{n}= \frac{n}{2} (2a+(n-1)dS

n

=

2

n

(2a+(n−1)d

\sf S_{n}=\frac{n}{2} (2(5)+(n-1)6S

n

=

2

n

(2(5)+(n−1)6

\sf⟹\frac{n}{2} (10+6n-6)⟹

2

n

(10+6n−6)

\sf ⟹S_{n} = \frac{n}{2} (6n+4)⟹S

n

=

2

n

(6n+4)

\sf ⟹S_{n}=\frac{6n²+4n}{2}⟹S

n

=

2

6n²+4n

⟹ Sn = 4n²+2n