Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Find the sum of nth terms of the positive perfect square integers, where,

Find the sum of nth terms of the positive perfect square integers, where, \mathrm{n=\sqrt{ } a} , whose nth term is \mathrm{(n) ! \times n}.

Option: 1

3


Option: 2

-2


Option: 3

-1


Option: 4

5


Answers (1)

best_answer

To find the sum of the nth terms of the positive perfect square integers, where n=\sqrt{ } aand a is a positive integer, with the nth term as (n) ! \times n,  we need to substitute the values of n=\sqrt{ } a into the given expression and then sum them up.
Let's consider a few values of a to find the sum of the nth terms:

\mathrm{For \, \mathrm{a}=1}:
\mathrm{n=\sqrt{ } 1=1}

For the 1st term \mathrm{(n=1)}:
\mathrm{(1) ! \times 1=1 \times 1=1}

\mathrm{For \, \mathrm{a}=4 :}
\mathrm{n=\sqrt{ } 4=2}

For the 2nd term \mathrm{(n=2):}
\mathrm{(2) ! \times 2=2 \times 2=4}
For a = 9:
\mathrm{n=\sqrt{9}=3}

For the 3rd term \mathrm{(n=3) :}
\mathrm{(3) ! \times 3=6 \times 3=18}
\mathrm{For \, \mathrm{a}=16 :}
\mathrm{n=\sqrt{ } 16=4}

For the 4 th term \mathrm{(n=4) :}
\mathrm{(4) ! \times 4=24 \times 4=96}
For a = 25:
\mathrm{n= \sqrt{25}= 5}

For the 5th term \mathrm{(n = 5):}
\mathrm{(5) ! \times 5=120 \times 5=600}
As we can see, the nth term is increasing as a increases. Therefore, we can observe that the nth term is given by \mathrm{n! \times n= \left ( n+1 \right )!-n!}

Now, let's calculate the sum of the nth terms:
\begin{aligned} & \text { Sum }=(1+1) !-1 !+(2+1) !-2 !+(3+1) !-3 !+(4+1) !-4 !+\ldots \\ & =2 !-1 !+3 !-2 !+4 !-3 !+5 !-4 !+\ldots \\ & =2+3 !-2+4 !-3+5 !-4+\ldots \\ & =2+(3 !-2 !)+(4 !-3 !)+(5 !-4 !)+\ldots \end{aligned}

We can observe that the common terms cancel out, leaving us with:
\mathrm{\text { Sum }=2+2 !+3 !+4 !+\ldots}

Now, let's calculate the sum of the terms:
\begin{aligned} & \text { Sum }=2+2+6+24+\ldots \\ & =2+2(1+3+12+\ldots) \\ & =2+2(1+3(1+4+\ldots)) \\ & =2+2(1+3(1+4(1+\ldots))) \end{aligned}

This is an infinite geometric series with the first term \mathrm{(a)=2} and the common ratio \mathrm{(r)=3}.
Using the formula for the sum of an infinite geometric series, we have:
\begin{aligned} & \text { Sum }=a /(1-r) \\ & =2 /(1-3) \\ & =2 /(-2) \\ & =-1 \end{aligned}

Therefore, the sum of the nth terms of the positive perfect square integers, where n=\sqrt[V]{ } a and a is a positive integer, with the nth term as n=\sqrt{ } a(n) ! \times n, is -1.


 


 


 

 

Posted by

Irshad Anwar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024: NTA debars 39 candidates for 3 years on account of using unfair means
anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

Latest Question