Given,Summation (r=1 to n ) Tr = n(n+1)(n+2)(n+3)/8 Find,Lt ( n tends to infinity )1/ Tr ?

To tackle the problem you've presented, we need to analyze the summation and its limit as \( n \) approaches infinity. The summation given is:

Summation (r=1 to n) Tr = n(n+1)(n+2)(n+3)/8

We want to find the limit:

Lt (n tends to infinity) 1/Tr

Breaking Down the Summation

First, let's understand what \( Tr \) represents in this context. The expression indicates that the sum of \( Tr \) from \( r = 1 \) to \( n \) equals \( \frac{n(n+1)(n+2)(n+3)}{8} \). To find \( Tr \), we need to express it in terms of \( n \).

Finding \( Tr \)

To find \( Tr \), we can use the formula for the sum of the first \( n \) terms of a polynomial. The expression \( \frac{n(n+1)(n+2)(n+3)}{8} \) is a polynomial of degree 4. To find \( Tr \), we can derive it from the summation formula. The \( r \)-th term can be approximated as:

Tr ≈ \frac{(n)(n+1)(n+2)(n+3)}{8} - \frac{(n-1)(n)(n+1)(n+2)}{8}

By simplifying this, we can find \( Tr \) for large \( n \). However, for our limit, we can also use the leading term of the polynomial to approximate \( Tr \) as \( n^4/32 \) for large \( n \).

Calculating the Limit

Now, we need to find:

Lt (n tends to infinity) 1/Tr

Substituting our approximation for \( Tr \), we have:

1/Tr ≈ 1/(n^4/32) = 32/n^4

As \( n \) approaches infinity, \( 32/n^4 \) approaches 0. Therefore:

Final Result

The limit can be expressed as:

Lt (n tends to infinity) 1/Tr = 0

This means that as \( n \) becomes very large, the value of \( 1/Tr \) approaches zero. This result is intuitive because as the terms in the summation grow larger, their reciprocals shrink towards zero.

In summary, through careful analysis of the polynomial and its behavior as \( n \) increases, we find that the limit of \( 1/Tr \) as \( n \) approaches infinity is indeed zero. This illustrates the concept of growth rates in sequences and series, where higher degree terms dominate the behavior of the function.