VN or Telescoping Method Questions in English

(A) The given series is $S_n = \sum_{r=1}^{n} \frac{1}{(2r)(2r+2)}$.
We can write the general term as $\frac{1}{4} \times \frac{1}{r(r+1)} = \frac{1}{4} \left( \frac{1}{r} - \frac{1}{r+1} \right)$.
Summing this from $r=1$ to $n$:
$S_n = \frac{1}{4} \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) \right]$.
This is a telescoping series,so $S_n = \frac{1}{4} \left( 1 - \frac{1}{n+1} \right)$.
$S_n = \frac{1}{4} \left( \frac{n+1-1}{n+1} \right) = \frac{n}{4(n+1)}$.
Comparing this with $\frac{k n}{n+1}$,we get $k = \frac{1}{4}$.

(D) Given $t_n = \frac{1}{4}(n+2)(n+3)$.
Then,$\frac{1}{t_n} = \frac{4}{(n+2)(n+3)}$.
Using partial fractions,$\frac{1}{t_n} = 4 \left[ \frac{1}{n+2} - \frac{1}{n+3} \right]$.
Let $S = \sum_{n=1}^{2003} \frac{1}{t_n} = 4 \sum_{n=1}^{2003} \left( \frac{1}{n+2} - \frac{1}{n+3} \right)$.
This is a telescoping series:
$S = 4 \left[ \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \ldots + \left( \frac{1}{2005} - \frac{1}{2006} \right) \right]$.
$S = 4 \left( \frac{1}{3} - \frac{1}{2006} \right)$.
$S = 4 \left( \frac{2006 - 3}{3 \times 2006} \right) = 4 \left( \frac{2003}{6018} \right) = \frac{8012}{6018} = \frac{4006}{3009}$.

(B) The given expression is $1000 \left[ \sum_{n=1}^{999} \frac{1}{n(n+1)} \right]$.
Using the method of partial fractions,we know that $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
Substituting this into the sum,we get $1000 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \ldots + (\frac{1}{999} - \frac{1}{1000}) \right]$.
This is a telescoping series where all intermediate terms cancel out,leaving $1000 \left[ 1 - \frac{1}{1000} \right]$.
$= 1000 \left[ \frac{999}{1000} \right] = 999$.

(B) The general term of the series is $T_n = n \times n!$.
We can rewrite $T_n$ as $T_n = ((n + 1) - 1) \times n! = (n + 1)! - n!$.
The sum $S_{50} = \sum_{n=1}^{50} T_n = \sum_{n=1}^{50} ((n + 1)! - n!)$.
This is a telescoping series:
$S_{50} = (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + (51! - 50!)$.
All intermediate terms cancel out,leaving $S_{50} = 51! - 1!$.
Since $1! = 1$,the sum is $51! - 1$.

(A) We have the term $T_r = \frac{r}{r^4+r^2+1}$.
Using the identity $r^4+r^2+1 = (r^2+r+1)(r^2-r+1)$,we can write:
$T_r = \frac{r}{(r^2+r+1)(r^2-r+1)} = \frac{1}{2} \left( \frac{1}{r^2-r+1} - \frac{1}{r^2+r+1} \right)$.
Let $f(r) = \frac{1}{r^2-r+1}$. Then $f(r+1) = \frac{1}{(r+1)^2-(r+1)+1} = \frac{1}{r^2+2r+1-r-1+1} = \frac{1}{r^2+r+1}$.
Thus,$T_r = \frac{1}{2} (f(r) - f(r+1))$.
The sum is $S = \sum_{r=1}^{25} T_r = \frac{1}{2} \sum_{r=1}^{25} (f(r) - f(r+1))$.
This is a telescoping sum: $S = \frac{1}{2} (f(1) - f(26))$.
$f(1) = \frac{1}{1^2-1+1} = 1$.
$f(26) = \frac{1}{26^2-26+1} = \frac{1}{676-26+1} = \frac{1}{651}$.
$S = \frac{1}{2} (1 - \frac{1}{651}) = \frac{1}{2} (\frac{650}{651}) = \frac{325}{651}$.
Since $\gcd(325, 651) = 1$,we have $p=325$ and $q=651$.
Therefore,$p+q = 325 + 651 = 976$.

(C) The general term $T_k = \frac{k}{1+4k^4}$.
We can rewrite the denominator as $1+4k^4 = 1+4k^4+4k^2-4k^2 = (2k^2+1)^2 - (2k)^2 = (2k^2-2k+1)(2k^2+2k+1)$.
Using partial fractions,$T_k = \frac{1}{4} \left[ \frac{1}{2k^2-2k+1} - \frac{1}{2k^2+2k+1} \right]$.
Let $f(k) = 2k^2-2k+1$. Then $f(k+1) = 2(k+1)^2-2(k+1)+1 = 2k^2+4k+2-2k-2+1 = 2k^2+2k+1$.
Thus,$T_k = \frac{1}{4} [\frac{1}{f(k)} - \frac{1}{f(k+1)}]$.
The sum of the first $10$ terms is $S_{10} = \sum_{k=1}^{10} T_k = \frac{1}{4} \sum_{k=1}^{10} [\frac{1}{f(k)} - \frac{1}{f(k+1)}]$.
This is a telescoping sum: $S_{10} = \frac{1}{4} [\frac{1}{f(1)} - \frac{1}{f(11)}]$.
$f(1) = 2(1)^2-2(1)+1 = 1$.
$f(11) = 2(11)^2-2(11)+1 = 242-22+1 = 221$.
$S_{10} = \frac{1}{4} [1 - \frac{1}{221}] = \frac{1}{4} \cdot \frac{220}{221} = \frac{55}{221}$.
Since $\text{gcd}(55, 221) = 1$,we have $m=55$ and $n=221$.
Therefore,$m+n = 55+221 = 276$.

(B) The general term is $T_n = 528 \cdot \frac{1}{n(n+1)(n+2)}$.
We use the partial fraction decomposition identity: $\frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$.
Thus,the sum is $\sum_{n=1}^{10} T_n = \frac{528}{2} \sum_{n=1}^{10} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$.
This is a telescoping series: $264 \left[ \left( \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} \right) + \left( \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} \right) + \dots + \left( \frac{1}{10 \cdot 11} - \frac{1}{11 \cdot 12} \right) \right]$.
All intermediate terms cancel out,leaving: $264 \left[ \frac{1}{2} - \frac{1}{132} \right]$.
$= 264 \left[ \frac{66 - 1}{132} \right] = 264 \cdot \frac{65}{132} = 2 \cdot 65 = 130$.