Question

find the sum of series root 3+ 3 root 2 +6 root 3 up to 16 terms

Answer 1

Given series,

$\sf \longrightarrow \sqrt{3} + 3 \sqrt{2} + 6 \sqrt{3} + ...$

We have to find the sum of the series upto 16 terms.

The given series can be rewritten as,

$\sf \longrightarrow \sqrt{3} + \sqrt{3} \sqrt{3} \sqrt{2} + \sqrt{6} \sqrt{6} \sqrt{3} + ...$

$\sf \longrightarrow \sqrt{3} + \sqrt{3} \sqrt{6} + \sqrt{6} \sqrt{6} \sqrt{3} + ...$

Now, we can see that this is a GP with first term √3 and common ratio √6.

Let's assume that,

• First term = a = √3
• Common ratio = r = √6

To find the sum of a GP upto nth terms whose first term is a and common ratio is r, we use the below formula:

$\boxed{ \sf S_n = \dfrac{a \left[r^{n} - 1 \right]}{r-1}}$

By substituting the known values in the formula, we get:

${ \implies \sf S_{16} = \dfrac{ \sqrt{3} \left[( \sqrt{6}) ^{16} - 1 \right]}{ \sqrt{6} -1}}$

${ \implies \sf S_{16} = \dfrac{ \sqrt{3} \left[( (\sqrt{6})^{2} ) ^{8} - 1 \right]}{ \sqrt{6} -1}}$

$\underline{ \implies \sf S_{16} = \dfrac{ \sqrt{3} \left[(6) ^{8} - 1 \right]}{ \sqrt{6} -1}}$

We can leave it here. This is the required answer.

Answer 2

Step-by-step explanation:

Given series,

\sf \longrightarrow \sqrt{3} + 3 \sqrt{2} + 6 \sqrt{3} + ...⟶

3

+3

2

+6

3

+...

We have to find the sum of the series upto 16 terms.

The given series can be rewritten as,

\sf \longrightarrow \sqrt{3} + \sqrt{3} \sqrt{3} \sqrt{2} + \sqrt{6} \sqrt{6} \sqrt{3} + ...⟶

3

+

3

3

2

+

6

6

3

+...

\sf \longrightarrow \sqrt{3} + \sqrt{3} \sqrt{6} + \sqrt{6} \sqrt{6} \sqrt{3} + ...⟶

3

+

3

6

+

6

6

3

+...

Now, we can see that this is a GP with first term √3 and common ratio √6.

Let's assume that,

First term = a = √3

Common ratio = r = √6

To find the sum of a GP upto nth terms whose first term is a and common ratio is r, we use the below formula:

\boxed{ \sf S_n = \dfrac{a \left[r^{n} - 1 \right]}{r-1}}

S

n

=

r−1

a[r

n

−1]

By substituting the known values in the formula, we get:

{ \implies \sf S_{16} = \dfrac{ \sqrt{3} \left[( \sqrt{6}) ^{16} - 1 \right]}{ \sqrt{6} -1}}⟹S

16

=

6

−1

3

[(

6

)

16

−1]

{ \implies \sf S_{16} = \dfrac{ \sqrt{3} \left[( (\sqrt{6})^{2} ) ^{8} - 1 \right]}{ \sqrt{6} -1}}⟹S

16

=

6

−1

3

[((

6

)

2

)

8

−1]

\underline{ \implies \sf S_{16} = \dfrac{ \sqrt{3} \left[(6) ^{8} - 1 \right]}{ \sqrt{6} -1}}

⟹S

16

=

6

−1

3

[(6)

8

−1]

We can leave it here. This is the required answer.