Answer

Verified

453.3k+ views

Hint: Here, we will be using the formulas for ${n^{{\text{th}}}}$ term and the sum of first $n$ terms of an AP.

Since, the odd integers occurring from 1 to 2001 are $1,3,5,.....,1997,1999,2001$.

Here, ${a_1} = 1$, $d = 2$ and ${a_n} = 2001$(last term)

Clearly, the above series have a common difference of 2 and represents an arithmetic progression.

For an AP having the first term as ${a_1}$ and common difference as $d$, ${n^{{\text{th}}}}$ term of the AP is given by ${a_n} = {a_1} + \left( {n - 1} \right)d$

For the given series, ${n^{{\text{th}}}}$term (last term) of the AP is given by $2001 = 1 + \left( {n - 1} \right) \times 2 \Rightarrow 2001 = 1 + 2n - 2 \Rightarrow 2n = 2002 \Rightarrow n = 1001$

Therefore, the total number of terms in the given series is 1001.

Also, the formula for the sum of first $n$ terms of the AP is given by ${{\text{S}}_n} = \dfrac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$

So, the required sum of the given series is given by ${{\text{S}}_n} = \dfrac{{1001}}{2}\left[ {2\left( 1 \right) + \left( {1001 - 1} \right) \times 2} \right] = \dfrac{{1001}}{2}\left[ {2 + 2000} \right] = \dfrac{{1001 \times 2002}}{2} = 1002001$

Note: The difference between any two consecutive odd integers and between any two consecutive even integers is two because in between any two consecutive odd integers, one even integer occurs and similarly in between any two consecutive even integers, one odd integer occurs. The important thing in such types of problems is to identify the type of progression.

Since, the odd integers occurring from 1 to 2001 are $1,3,5,.....,1997,1999,2001$.

Here, ${a_1} = 1$, $d = 2$ and ${a_n} = 2001$(last term)

Clearly, the above series have a common difference of 2 and represents an arithmetic progression.

For an AP having the first term as ${a_1}$ and common difference as $d$, ${n^{{\text{th}}}}$ term of the AP is given by ${a_n} = {a_1} + \left( {n - 1} \right)d$

For the given series, ${n^{{\text{th}}}}$term (last term) of the AP is given by $2001 = 1 + \left( {n - 1} \right) \times 2 \Rightarrow 2001 = 1 + 2n - 2 \Rightarrow 2n = 2002 \Rightarrow n = 1001$

Therefore, the total number of terms in the given series is 1001.

Also, the formula for the sum of first $n$ terms of the AP is given by ${{\text{S}}_n} = \dfrac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$

So, the required sum of the given series is given by ${{\text{S}}_n} = \dfrac{{1001}}{2}\left[ {2\left( 1 \right) + \left( {1001 - 1} \right) \times 2} \right] = \dfrac{{1001}}{2}\left[ {2 + 2000} \right] = \dfrac{{1001 \times 2002}}{2} = 1002001$

Note: The difference between any two consecutive odd integers and between any two consecutive even integers is two because in between any two consecutive odd integers, one even integer occurs and similarly in between any two consecutive even integers, one odd integer occurs. The important thing in such types of problems is to identify the type of progression.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How many crores make 10 million class 7 maths CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE