Question

please solve question number 12 ​

Answer 1

Question:-

➡ Find the sum of the series upto 3n terms.

Solution:-

Given series,

$\sf S = \red3 + \blue5 + \orange7 + \red6 + \blue9 + \orange{12} + \red9 + \blue{13}+ \orange{17}+ ...3n$

We have to derive a formula to calculate the sum of the series.

From above, we can see that this infinite series is a combination of three series,

$\sf S_{1} = \red{3 + 6 + 9 + ...n}$

$\sf S_{2} = \blue{5 + 9 + 13 + ...n}$

$\sf S_{3} = \orange{7 + 12 + 17+ ...n}$

$\sf \implies S = S_{1} + S_{2} + S_{3}$

Now, we will calculate sum of first n terms of the three series.

In $\sf S_{1}$,

➡ a = 3 (first term)

➡ d = 6 - 3 = 3 (common difference)

Formula to calculate sum of first nth term of AP is,

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

So,

$\sf S_{n} = \frac{n}{2} \{2 \times 3+ (n - 1) \times 3 \}$

$\sf \implies S_{n} = \frac{n}{2} \{6+3n - 3\}$

$\sf \implies S_{n} = \frac{n}{2} \{3n + 3 \} \: ...(i)$

Now, in $\sf S_{2}$,

➡ a = 5 (first term)

➡ d = 9 - 5 = 4 (common difference)

So,

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

$\sf \implies S_{n} = \frac{n}{2} \{2 \times 5+ (n - 1) \times 4 \}$

$\sf \implies S_{n} = \frac{n}{2} \{10+ 4n - 4\}$

$\sf \implies S_{n} = \frac{n}{2} \{ 4n + 6\} \: ...(ii)$

Now, in $\sf S_{3}$

➡ a = 7 (first term)

➡ d = 12 - 7 = 5 (common difference)

So,

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

$\sf \implies S_{n} = \frac{n}{2} \{2 \times 7+ (n - 1) \times 5 \}$

$\sf \implies S_{n} = \frac{n}{2} \{14+ 5n - 5\}$

$\sf \implies S_{n} = \frac{n}{2} \{5n + 9\} \: ...(iii)$

Now, adding equations (i), (ii) and (iii), we get,

$\sf S = \frac{n}{2} \{ (3n + 3 )+ (4n + 6) + (5n + 9)\}$

$\sf \implies S = \frac{n}{2} \{12n + 18\}$

$\sf \implies S = n(6n + 9)$

$\sf \implies S = 3n(2n + 3)$

Hence, the sum of the series upto 3n terms is 3n(2n + 3)

Answer:-

➡ The sum of the series upto 3n terms is 3n(2n + 3)

Answer 2

Answer:

Given series,

\sf S = \red3 + \blue5 + \orange7 + \red6 + \blue9 + \orange{12} + \red9 + \blue{13}+ \orange{17}+ ...3nS=3+5+7+6+9+12+9+13+17+...3n

We have to derive a formula to calculate the sum of the series.

From above, we can see that this infinite series is a combination of three series,

\sf S_{1} = \red{3 + 6 + 9 + ...n}S

1

=3+6+9+...n

\sf S_{2} = \blue{5 + 9 + 13 + ...n}S

2

=5+9+13+...n

\sf S_{3} = \orange{7 + 12 + 17+ ...n}S

3

=7+12+17+...n

\sf \implies S = S_{1} + S_{2} + S_{3}⟹S=S

1

+S

2

+S

3

Now, we will calculate sum of first n terms of the three series.

In \sf S_{1}S

1

,

➡ a = 3 (first term)

➡ d = 6 - 3 = 3 (common difference)

Formula to calculate sum of first nth term of AP is,

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

n

=

2

n

{2a+(n−1)d}

So,

\sf S_{n} = \frac{n}{2} \{2 \times 3+ (n - 1) \times 3 \}S

n

=

2

n

{2×3+(n−1)×3}

\sf \implies S_{n} = \frac{n}{2} \{6+3n - 3\}⟹S

n

=

2

n

{6+3n−3}

\sf \implies S_{n} = \frac{n}{2} \{3n + 3 \} \: ...(i)⟹S

n

=

2

n

{3n+3}...(i)

Now, in \sf S_{2}S

2

,

➡ a = 5 (first term)

➡ d = 9 - 5 = 4 (common difference)

So,

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

n

=

2

n

{2a+(n−1)d}

\sf \implies S_{n} = \frac{n}{2} \{2 \times 5+ (n - 1) \times 4 \}⟹S

n

=

2

n

{2×5+(n−1)×4}

\sf \implies S_{n} = \frac{n}{2} \{10+ 4n - 4\}⟹S

n

=

2

n

{10+4n−4}

\sf \implies S_{n} = \frac{n}{2} \{ 4n + 6\} \: ...(ii)⟹S

n

=

2

n

{4n+6}...(ii)

Now, in \sf S_{3}S

3

➡ a = 7 (first term)

➡ d = 12 - 7 = 5 (common difference)

So,

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

n

=

2

n

{2a+(n−1)d}

\sf \implies S_{n} = \frac{n}{2} \{2 \times 7+ (n - 1) \times 5 \}⟹S

n

=

2

n

{2×7+(n−1)×5}

\sf \implies S_{n} = \frac{n}{2} \{14+ 5n - 5\}⟹S

n

=

2

n

{14+5n−5}

\sf \implies S_{n} = \frac{n}{2} \{5n + 9\} \: ...(iii)⟹S

n

=

2

n

{5n+9}...(iii)

Now, adding equations (i), (ii) and (iii), we get,

\sf S = \frac{n}{2} \{ (3n + 3 )+ (4n + 6) + (5n + 9)\}S=

2

n

{(3n+3)+(4n+6)+(5n+9)}

\sf \implies S = \frac{n}{2} \{12n + 18\}⟹S=

2

n

{12n+18}

\sf \implies S = n(6n + 9)⟹S=n(6n+9)

\sf \implies S = 3n(2n + 3)⟹S=3n(2n+3)

Hence, the sum of the series upto 3n terms is 3n(2n + 3)

Answer:-

➡ The sum of the series upto 3n terms is 3n(2n + 3)

Step-by-step explanation:

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Warm regards:Miss Chikchiki