Key Concepts and Formulas

• Sum of first $n$ natural numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
• Sum of squares of first $n$ natural numbers: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
• Sum of cubes of first $n$ natural numbers: $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$

Step-by-Step Solution

Step 1: Express the given expression using summation notation. The given expression is a ratio of two series. We can represent the general term of each series and then use summation notation. The numerator is $1 \times 2^2 + 2 \times 3^2 + \ldots + 100 \times (101)^2$. The $n$-th term is $n(n+1)^2$. So, the numerator is $\sum_{n=1}^{100} n(n+1)^2$. The denominator is $1^2 \times 2 + 2^2 \times 3 + \ldots + 100^2 \times 101$. The $n$-th term is $n^2(n+1)$. So, the denominator is $\sum_{n=1}^{100} n^2(n+1)$. The expression becomes: $\frac{\sum_{n=1}^{100} n(n+1)^2}{\sum_{n=1}^{100} n^2(n+1)}$

Step 2: Expand the terms within the summations. We expand the general terms to express them as sums of powers of $n$. For the numerator: $n(n+1)^2 = n(n^2 + 2n + 1) = n^3 + 2n^2 + n$ So, the numerator sum is: $\sum_{n=1}^{100} (n^3 + 2n^2 + n) = \sum_{n=1}^{100} n^3 + 2\sum_{n=1}^{100} n^2 + \sum_{n=1}^{100} n$ For the denominator: $n^2(n+1) = n^3 + n^2$ So, the denominator sum is: $\sum_{n=1}^{100} (n^3 + n^2) = \sum_{n=1}^{100} n^3 + \sum_{n=1}^{100} n^2$

Step 3: Apply the summation formulas. Now we substitute the formulas for the sum of cubes, squares, and natural numbers with $n=100$. Let $S_3 = \sum_{n=1}^{100} n^3$, $S_2 = \sum_{n=1}^{100} n^2$, and $S_1 = \sum_{n=1}^{100} n$. $S_1 = \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050$. $S_2 = \frac{100(100+1)(2 \times 100+1)}{6} = \frac{100 \times 101 \times 201}{6} = 50 \times 101 \times \frac{201}{3} = 50 \times 101 \times 67$. $S_3 = \left(\frac{100(100+1)}{2}\right)^2 = (5050)^2$.

The numerator sum is $S_3 + 2S_2 + S_1$. The denominator sum is $S_3 + S_2$.

Step 4: Substitute the expanded sums back into the fraction and simplify. The expression becomes: $\frac{S_3 + 2S_2 + S_1}{S_3 + S_2}$ We can rewrite this by separating terms: $\frac{(S_3 + S_2) + S_2 + S_1}{S_3 + S_2} = 1 + \frac{S_2 + S_1}{S_3 + S_2}$ Let's substitute the values or expressions for $S_1, S_2, S_3$. It's often easier to work with the expressions before calculating large numbers. Numerator sum: $\frac{100(101)(201)}{6} + \frac{100(101)}{2} + \left(\frac{100(101)}{2}\right)^2$ Denominator sum: $\frac{100(101)(201)}{6} + \left(\frac{100(101)}{2}\right)^2$

Let $X = \frac{100(101)}{2} = 5050$ and $Y = \frac{100(101)(201)}{6}$. The expression is $\frac{Y + 2 \frac{X}{5050} X + X}{Y + X} = \frac{Y + 2S_2 + S_1}{Y + S_2}$. This substitution is not simplifying much.

Let's go back to the expanded sums: Numerator: $\sum_{n=1}^{100} n^3 + 2\sum_{n=1}^{100} n^2 + \sum_{n=1}^{100} n$ Denominator: $\sum_{n=1}^{100} n^3 + \sum_{n=1}^{100} n^2$

We can factor out common terms. Let $N=100$. Numerator: $\frac{N^2(N+1)^2}{4} + 2 \frac{N(N+1)(2N+1)}{6} + \frac{N(N+1)}{2}$ Denominator: $\frac{N^2(N+1)^2}{4} + \frac{N(N+1)(2N+1)}{6}$

Let $A = \frac{N(N+1)}{2}$ and $B = \frac{N(N+1)(2N+1)}{6}$. Numerator: $A^2 + 2B + A$ Denominator: $A^2 + B$

The expression is $\frac{A^2 + 2B + A}{A^2 + B}$. Substitute $N=100$: $A = \frac{100 \times 101}{2} = 5050$. $B = \frac{100 \times 101 \times 201}{6} = 50 \times 101 \times 67 = 338350$. $A^2 = (5050)^2 = 25502500$.

Numerator = $25502500 + 2(338350) + 5050 = 25502500 + 676700 + 5050 = 26184250$. Denominator = $25502500 + 338350 = 25840850$.

The fraction is $\frac{26184250}{25840850}$. Divide by 10: $\frac{2618425}{2584085}$. Both are divisible by 5. $2618425 / 5 = 523685$. $2584085 / 5 = 516817$. So we have $\frac{523685}{516817}$.

Let's try simplifying the expression in terms of $N$ before substituting. Numerator: $\frac{N^2(N+1)^2}{4} + \frac{N(N+1)(2N+1)}{3} + \frac{N(N+1)}{2}$ Common factor $N(N+1)$: $N(N+1) \left[ \frac{N(N+1)}{4} + \frac{2N+1}{3} + \frac{1}{2} \right]$ $= N(N+1) \left[ \frac{3N(N+1) + 4(2N+1) + 6}{12} \right]$ $= N(N+1) \left[ \frac{3N^2+3N + 8N+4 + 6}{12} \right]$ $= N(N+1) \left[ \frac{3N^2+11N+10}{12} \right]$ Factor the quadratic $3N^2+11N+10$: $(3N+5)(N+2)$. Numerator = $\frac{N(N+1)(3N+5)(N+2)}{12}$.

Denominator: $\frac{N^2(N+1)^2}{4} + \frac{N(N+1)(2N+1)}{6}$ Common factor $N(N+1)$: $N(N+1) \left[ \frac{N(N+1)}{4} + \frac{2N+1}{6} \right]$ $= N(N+1) \left[ \frac{3N(N+1) + 2(2N+1)}{12} \right]$ $= N(N+1) \left[ \frac{3N^2+3N + 4N+2}{12} \right]$ $= N(N+1) \left[ \frac{3N^2+7N+2}{12} \right]$ Factor the quadratic $3N^2+7N+2$: $(3N+1)(N+2)$. Denominator = $\frac{N(N+1)(3N+1)(N+2)}{12}$.

Now, the ratio is: $\frac{\frac{N(N+1)(3N+5)(N+2)}{12}}{\frac{N(N+1)(3N+1)(N+2)}{12}} = \frac{3N+5}{3N+1}$

Step 5: Substitute $N=100$ into the simplified expression. $\frac{3(100)+5}{3(100)+1} = \frac{300+5}{300+1} = \frac{305}{301}$

Common Mistakes & Tips

• Algebraic Errors: Expanding and simplifying algebraic expressions can lead to errors. It's beneficial to factor out common terms as early as possible.
• Formula Recall: Ensure the correct formulas for sums of powers are memorized.
• Substitution Strategy: Sometimes it's better to simplify the expression algebraically in terms of $N$ before substituting the numerical value of $N$, as this reduces the chance of arithmetic errors with large numbers.

Summary The problem involves calculating the ratio of two series. By expressing these series using summation notation and then expanding the general terms, we obtained sums of powers of $n$. Applying the standard formulas for the sum of cubes, squares, and natural numbers, and then carefully simplifying the resulting algebraic expression in terms of $N$, we arrived at a simple ratio. Substituting $N=100$ into this ratio gives the final answer.

The final answer is $\boxed{\frac{305}{301}}$ which corresponds to option (A).