Question

The value of $\frac{1\times 2^2+2\times3^2+...+100\times (101)^2}{1^2 \times 2\times2^2 \times3+...+100^2 \times 101}$

• A. 305/301
• B. 301/305
• C. 350/310
• D. 310/350

Answer 1

The Correct Option is A

To solve the problem, we need to find the value of the expression:

\(\frac{1 \times 2^2 + 2 \times 3^2 + \ldots + 100 \times (101)^2}{1^2 \times 2 \times 2^2 \times 3 + \ldots + 100^2 \times 101}\) 

Let's break this down step by step.

1. The numerator can be written as a sum of terms, where each term is \(n \times (n+1)^2\):
2. Similarly, the denominator is a sum of terms, where each term is \(n^2 \times (n+1)\):
3. Both series represent polynomial expansions:
   • \(n \times (n+1)^2 = n \times (n^2 + 2n + 1) = n^3 + 2n^2 + n\)
   • \(n^2 \times (n+1) = n^3 + n^2\)
4. We observe that the terms in the numerator and the denominator can be rearranged as follows:
   • For the numerator: \(n^3 + 2n^2 + n\)
   • For the denominator: \(n^3 + n^2\)
5. Now let's look at the ratio of a single term from both parts:
   • For each term: \(\frac{n^3 + 2n^2 + n}{n^3 + n^2} = \frac{n(n^2 + 2n + 1)}{n^2(n+1)} = \frac{n(n+1)^2}{n^2(n+1)}\)
   • This simplifies by cancelling common terms: \(= \frac{n+1}{n}\)
6. Since the ratio simplifies uniformly across the terms, sum of each simplified term's contribution:
   • \(= \frac{\sum_{n=1}^{100} (n+1)}{\sum_{n=1}^{100} n}\)
   • The numerator, \(\sum_{n=1}^{100} (n+1) = 1 + 2 + \cdots + 101\) which is a sum of an arithmetic progression with 101 terms.
   • The denominator is \(\sum_{n=1}^{100} n\), an arithmetic series with 100 terms.
7. Simplify the progression:
   • \(1 + 2 + \cdots + 101 = \frac{101 \times 102}{2} = 5151\)
   • \(1 + 2 + \cdots + 100 = \frac{100 \times 101}{2} = 5050\)
8. Thus, the expression becomes:
9. Calculate the expression:
10. Converting this simplified fraction to the options provided, the answer is:
    • \(\frac{305}{301}\) which is approximately \(1.013\)
    • Thus, the closest fraction matching our calculation is \(\frac{305}{301}\)

The correct answer is \(\frac{305}{301}\)