Math Problem Statement

Ans

Let S_n = sum from r=1 to n of (n^2 + nr + r^2) / n^3 and T_n = sum from r=0 to n-1 of (n^2 + nr + r^2) / n^3 for n = 1, 2, 3, then...

Solution

The problem involves two sums, SnS_n and TnT_n, defined as follows:

Sn=r=1nn2+nr+r2n3S_n = \sum_{r=1}^{n} \frac{n^2 + nr + r^2}{n^3} and Tn=r=0n1n2+nr+r2n3T_n = \sum_{r=0}^{n-1} \frac{n^2 + nr + r^2}{n^3}

We are asked to determine whether TnT_n or SnS_n are greater or less than 116\frac{11}{6}.

Step-by-step Solution:

1. Simplification of SnS_n:

We begin by analyzing SnS_n:

Sn=r=1nn2+nr+r2n3S_n = \sum_{r=1}^{n} \frac{n^2 + nr + r^2}{n^3} Simplifying the terms inside the sum:

Sn=r=1n(n2n3+nrn3+r2n3)S_n = \sum_{r=1}^{n} \left( \frac{n^2}{n^3} + \frac{nr}{n^3} + \frac{r^2}{n^3} \right) This becomes:

Sn=r=1n(1n+rn2+r2n3)S_n = \sum_{r=1}^{n} \left( \frac{1}{n} + \frac{r}{n^2} + \frac{r^2}{n^3} \right) Now, split the sum:

Sn=1nr=1n1+1n2r=1nr+1n3r=1nr2S_n = \frac{1}{n} \sum_{r=1}^{n} 1 + \frac{1}{n^2} \sum_{r=1}^{n} r + \frac{1}{n^3} \sum_{r=1}^{n} r^2

Each of these sums can be evaluated:

  • r=1n1=n\sum_{r=1}^{n} 1 = n,
  • r=1nr=n(n+1)2\sum_{r=1}^{n} r = \frac{n(n+1)}{2},
  • r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}.

Substitute these values back into the equation for SnS_n:

Sn=nn+1n2n(n+1)2+1n3n(n+1)(2n+1)6S_n = \frac{n}{n} + \frac{1}{n^2} \cdot \frac{n(n+1)}{2} + \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} Simplify:

Sn=1+n+12n+(n+1)(2n+1)6n2S_n = 1 + \frac{n+1}{2n} + \frac{(n+1)(2n+1)}{6n^2}

2. Simplification of TnT_n:

Similarly, for TnT_n:

Tn=r=0n1n2+nr+r2n3T_n = \sum_{r=0}^{n-1} \frac{n^2 + nr + r^2}{n^3} Following the same steps, we arrive at:

Tn=1+n12n+(n1)(2n1)6n2T_n = 1 + \frac{n-1}{2n} + \frac{(n-1)(2n-1)}{6n^2}

3. Analysis:

For large values of nn, both SnS_n and TnT_n approach a limit. Comparing these to 116\frac{11}{6}, we find that:

  • SnS_n is slightly greater than 116\frac{11}{6} for large nn,
  • TnT_n is slightly less than 116\frac{11}{6} for large nn.

Thus, the correct answers are:

  • (A) Tn<116T_n < \frac{11}{6}
  • (D) Sn>116S_n > \frac{11}{6}

Conclusion:

The correct options are:

  • (A) Tn<116T_n < \frac{11}{6}
  • (D) Sn>116S_n > \frac{11}{6}

Would you like further clarification on any step or more details on specific aspects of the sums? Here are some related questions to explore:

  1. What are other methods to evaluate large summations like SnS_n and TnT_n?
  2. How does the sum r2\sum r^2 grow in comparison to r\sum r?
  3. Can we approximate SnS_n and TnT_n for small values of nn?
  4. How would the sums change if we used a different polynomial inside the summation?
  5. How do SnS_n and TnT_n behave as nn \to \infty?

Tip: When simplifying summations, always check if common summation formulas can be applied to avoid manual computation of terms.

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Math Problem Analysis

Mathematical Concepts

Summation
Series
Algebraic simplification

Formulas

S_n = sum from r=1 to n of (n^2 + nr + r^2) / n^3
T_n = sum from r=0 to n-1 of (n^2 + nr + r^2) / n^3
Sum of first n natural numbers: sum r = n(n+1)/2
Sum of squares of first n natural numbers: sum r^2 = n(n+1)(2n+1)/6

Theorems

Summation formulas for natural numbers and their squares
Limits of sequences for large n

Suitable Grade Level

Grades 11-12, Pre-College