Math Problem Statement

no 5

The fifth series provided is: 1/(2*3) + 1/(3*4) + 1/(4*5) + 1/(5*6) + ... + 1/(n+1)(n+2) + ...

Solution

The fifth series provided is:

123+134+145+156++1(n+1)(n+2)+\frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \frac{1}{5 \cdot 6} + \dots + \frac{1}{(n+1)(n+2)} + \dots

This series can be solved using partial fraction decomposition, just like we did before. Let's go through the process again to find its nth partial sum and the sum of the series.

Step 1: General term of the series.

The general term of the series is:

Tn=1(n+1)(n+2)T_n = \frac{1}{(n+1)(n+2)}

Step 2: Partial fraction decomposition.

We can decompose this term as:

1(n+1)(n+2)=An+1+Bn+2\frac{1}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2}

Multiplying both sides by (n+1)(n+2)(n+1)(n+2) to clear the denominators:

1=A(n+2)+B(n+1)1 = A(n+2) + B(n+1)

Expanding both sides:

1=A(n)+2A+B(n)+B1 = A(n) + 2A + B(n) + B 1=(A+B)n+(2A+B)1 = (A + B)n + (2A + B)

Equating the coefficients of nn and the constant term:

  • A+B=0A + B = 0
  • 2A+B=12A + B = 1

From A+B=0A + B = 0, we get B=AB = -A. Substituting this into the second equation:

2AA=1A=12A - A = 1 \quad \Rightarrow \quad A = 1

Thus, (B = -1

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Math Problem Analysis

Mathematical Concepts

Partial Fractions
Series
Summation

Formulas

Partial Fraction Decomposition
nth Partial Sum Formula for Series

Theorems

Partial Fraction Decomposition Theorem

Suitable Grade Level

Grades 11-12