Question

Find the sum of n terms of the series
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​

Answer 1

★ Concept :-

Here the concept of sequence and series has been used. We see that we are given a series. Firstly we can understand the pattern of the series and then find the nth term. Then after that we can find the sum of n terms of the sequence.

Let's do it !!

______________________________________

★ Formula Used :-

$\;\boxed{\sf{\pink{a_{n}\;=\;\bf{n(n\:+\:1)}}}}$

$\;\displaystyle{\boxed{\sf{\pink{S_{n}\;=\;\bf{\sum_{n\:=\:1} ^{n}\:a_{n}}}}}}$

______________________________________

★ Solution :-

Given sequence,

→ 1 × 2 + 2 × 3 + 3 × 4 + ...

Clearly we see that here one term multiplies itself with sum of 1. This means n × (n + 1)

• Since, 1 × 2 = 1 × (1 + 1) where n = 1

• 2 × 3 = 2 × (2 + 1) where n = 2

• 3 × 4 = 3 × (3 + 1) where n = 3

Now this series will continue further.

So pattern of series is n × (n + 1).

• Let the nth term of A.P. be aₙ

• Let the sum of n terms of A.P. be Sₙ

-----------------------------------------------------------

~ For the value of aₙ ::

We know that,

$\;\sf{\rightarrow\;\;a_{n}\;=\;\bf{n(n\:+\:1)}}$

Then since n = n only, so

$\;\bf{\rightarrow\;\;\red{a_{n}\;=\;\bf{n^{2}\:+\:n}}}$

-----------------------------------------------------------

~ For the value of Sₙ ::

We know that,

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\sum_{n\:=\:1} ^{n}\:a_{n}}}}$

We already have the value of aₙ so by applying that, we get

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\sum_{n\:=\:1} ^{n}\:(n^{2}\:+\:n)}}}$

Since Summation is distributive, so

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\sum_{n\:=\:1} ^{n}\:n^{2}\;+\;\sum_{n\:=\:1} ^{n}n}}}$

Now on calculating by using limits, we get

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)(2n\:+\:1)}{6}\;+\;\dfrac{n(n\:+\:1)}{2}}}}$

This can be written as ,

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{2}\:\times\:\dfrac{(2n\:+\:1)}{3}\;+\;\dfrac{n(n\:+\:1)}{2}}}}$

Now taking the common term, we get

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{2}\bigg(\dfrac{(2n\:+\:1)}{3}\;+\;1\bigg)}}}$

Now let's first solve the terms in bracket first.

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{2}\bigg(\dfrac{(2n\:+\:1)\:+\:3}{3}\bigg)}}}$

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{2}\bigg(\dfrac{2n\:+\:1\:+\:3}{3}\bigg)}}}$

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{2}\bigg(\dfrac{2n\:+\:4}{3}\bigg)}}}$

Now taking the denominator as single term, we get

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{6}(2n\:+\:4)}}}$

• Since 2 × 3 = 6

The above equation can be written as ,

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{6}[2(n\:+\:2)]}}}$

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{2n(n\:+\:1)}{6}(n\:+\:2)}}}$

$\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{2n(n\:+\:1)(n\:+\:2)}{6}}}}$

$\;\displaystyle{\sf{\rightarrow\;\;\green{S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)(n\:+\:2)}{3}}}}}$

This is the required answer.

$\;\underline{\boxed{\tt{Required\;\:Sum\;=\;\bf{\purple{\dfrac{n(n\:+\:1)(n\:+\:2)}{3}}}}}}$

________________________________

★ More to know :-

• For A.P. :-

$\;\tt{\leadsto\;\;Common\; Difference\;=\;a_{2}\;-\;a_{1}}$

• where a is the number of terms

$\;\tt{\leadsto\;\;a_{n}\;=\;a\;+\;(n\:-\:2)d}$

$\;\tt{\leadsto\;\;l_{n}\;=\;a\;+\;(n\:-\:2)(-d)}$

$\;\tt{\leadsto\;\;S_{n}\;=\;\dfrac{n}{2}[2a\:+\:(n\:-\;1)d]}$

$\;\tt{\leadsto\;\;S_{n}\;=\;\dfrac{n}{2}[a\;+\;a_{n}]}$

Answer 2

★ Concept :-

Here the concept of sequence and series has been used. We see that we are given a series. Firstly we can understand the pattern of the series and then find the nth term. Then after that we can find the sum of n terms of the sequence.

Let's do it !!

______________________________________

★ Formula Used :-

\;\boxed{\sf{\pink{a_{n}\;=\;\bf{n(n\:+\:1)}}}}

a

n

=n(n+1)

\;\displaystyle{\boxed{\sf{\pink{S_{n}\;=\;\bf{\sum_{n\:=\:1} ^{n}\:a_{n}}}}}}

S

n

=

n=1

∑

n

a

n

______________________________________

★ Solution :-

Given sequence,

→ 1 × 2 + 2 × 3 + 3 × 4 + ...

Clearly we see that here one term multiplies itself with sum of 1. This means n × (n + 1)

Since, 1 × 2 = 1 × (1 + 1) where n = 1

2 × 3 = 2 × (2 + 1) where n = 2

3 × 4 = 3 × (3 + 1) where n = 3

Now this series will continue further.

So pattern of series is n × (n + 1).

Let the nth term of A.P. be aₙ

Let the sum of n terms of A.P. be Sₙ

-----------------------------------------------------------

~ For the value of aₙ ::

We know that,

\;\sf{\rightarrow\;\;a_{n}\;=\;\bf{n(n\:+\:1)}}→a

n

=n(n+1)

Then since n = n only, so

\;\bf{\rightarrow\;\;\red{a_{n}\;=\;\bf{n^{2}\:+\:n}}}→a

n

=n

2

+n

-----------------------------------------------------------

~ For the value of Sₙ ::

We know that,

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\sum_{n\:=\:1} ^{n}\:a_{n}}}}→S

n

=

n=1

∑

n

a

n

We already have the value of aₙ so by applying that, we get

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\sum_{n\:=\:1} ^{n}\:(n^{2}\:+\:n)}}}→S

n

=

n=1

∑

n

(n

2

+n)

Since Summation is distributive, so

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\sum_{n\:=\:1} ^{n}\:n^{2}\;+\;\sum_{n\:=\:1} ^{n}n}}}→S

n

=

n=1

∑

n

n

2

+

n=1

∑

n

n

Now on calculating by using limits, we get

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)(2n\:+\:1)}{6}\;+\;\dfrac{n(n\:+\:1)}{2}}}}→S

n

=

6

n(n+1)(2n+1)

+

2

n(n+1)

This can be written as ,

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{2}\:\times\:\dfrac{(2n\:+\:1)}{3}\;+\;\dfrac{n(n\:+\:1)}{2}}}}→S

n

=

2

n(n+1)

×

3

(2n+1)

+

2

n(n+1)

Now taking the common term, we get

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{2}\bigg(\dfrac{(2n\:+\:1)}{3}\;+\;1\bigg)}}}→S

n

=

2

n(n+1)

(

3

(2n+1)

+1)

Now let's first solve the terms in bracket first.

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{2}\bigg(\dfrac{(2n\:+\:1)\:+\:3}{3}\bigg)}}}→S

n

=

2

n(n+1)

(

3

(2n+1)+3

)

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{2}\bigg(\dfrac{2n\:+\:1\:+\:3}{3}\bigg)}}}→S

n

=

2

n(n+1)

(

3

2n+1+3

)

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{2}\bigg(\dfrac{2n\:+\:4}{3}\bigg)}}}→S

n

=

2

n(n+1)

(

3

2n+4

)

Now taking the denominator as single term, we get

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{6}(2n\:+\:4)}}}→S

n

=

6

n(n+1)

(2n+4)

• Since 2 × 3 = 6

The above equation can be written as ,

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)}{6}[2(n\:+\:2)]}}}→S

n

=

6

n(n+1)

[2(n+2)]

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{2n(n\:+\:1)}{6}(n\:+\:2)}}}→S

n

=

6

2n(n+1)

(n+2)

\;\displaystyle{\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{2n(n\:+\:1)(n\:+\:2)}{6}}}}→S

n

=

6

2n(n+1)(n+2)

\;\displaystyle{\sf{\rightarrow\;\;\green{S_{n}\;=\;\bf{\dfrac{n(n\:+\:1)(n\:+\:2)}{3}}}}}→S

n

=

3

n(n+1)(n+2)

This is the required answer.

\;\underline{\boxed{\tt{Required\;\:Sum\;=\;\bf{\purple{\dfrac{n(n\:+\:1)(n\:+\:2)}{3}}}}}}

RequiredSum=

3

n(n+1)(n+2)

________________________________

★ More to know :-

• For A.P. :-

\;\tt{\leadsto\;\;Common\; Difference\;=\;a_{2}\;-\;a_{1}}⇝CommonDifference=a

2

−a

1

where a is the number of terms

\;\tt{\leadsto\;\;a_{n}\;=\;a\;+\;(n\:-\:2)d}⇝a

n

=a+(n−2)d

\;\tt{\leadsto\;\;l_{n}\;=\;a\;+\;(n\:-\:2)(-d)}⇝l

n

=a+(n−2)(−d)

\;\tt{\leadsto\;\;S_{n}\;=\;\dfrac{n}{2}[2a\:+\:(n\:-\;1)d]}⇝S

n

=

2

n

[2a+(n−1)d]

\;\tt{\leadsto\;\;S_{n}\;=\;\dfrac{n}{2}[a\;+\;a_{n}]}⇝S

n

=

2

n

[a+a

n

]