Verified answer

★ 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 :-

______________________________________

★ 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,

Then since n = n only, so

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

~ For the value of Sₙ ::

We know that,

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

Since Summation is distributive, so

Now on calculating by using limits, we get

This can be written as ,

Now taking the common term, we get

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

Now taking the denominator as single term, we get

• Since 2 × 3 = 6

The above equation can be written as ,

This is the required answer.

________________________________

★ More to know :-

• For A.P. :-

• where a is the number of terms

★ 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

]