What is the sum of series?? What is the sum of first 50 terms of the series
$$(1\times3)+(3\times5)+(5\times7)+\ldots$$
I had tried to solve this question 
It seems that it is mixed arithmetic series
In which first $A.p$ is $1,3,5\ldots$
Nd other one is $3,5,7\ldots$
But i don't know how to solve it together
Please give me a better solution or approach to solve this problem? 
Tell me a best way to solve 
Such problems 

You have better go through Falling Factorials
$$
\eqalign{
  & \sum\limits_{n = 0}^{49} {\left( {2n + 1} \right)\left( {2n + 3} \right)}  = 4\sum\limits_{n = 0}^{49} {\left( {n + 1/2} \right)\left( {n + 3/2} \right)}  =   \cr 
  &  = 4\sum\limits_{n = 0}^{49} {\left( {n + 3/2} \right)^{\,\underline {\,2\,} } }  \cr} 
$$
because then the sum telescopes nicely.   
In fact we have 
$$
\eqalign{
  & \left( {n + 3/2} \right)^{\,\underline {\,2\,} }  = {1 \over 3}\left( {\left( {n + 1 + 3/2} \right)^{\,\underline {\,3\,} }  - \left( {n + 3/2} \right)^{\,\underline {\,3\,} } } \right) =   \cr 
  &  = {1 \over 3}\left( {\left( {n + 1 + 3/2} \right)\left( {n + 3/2} \right)\left( {n - 1 + 3/2} \right)
 - \left( {n + 3/2} \right)\left( {n - 1 + 3/2} \right)\left( {n - 2 + 3/2} \right)} \right) =   \cr 
  &  = {1 \over 3}\left( {\left( {n + 1 + 3/2} \right) - \left( {n - 2 + 3/2} \right)} \right)\left( {\left( {n + 3/2} \right)\left( {n - 1 + 3/2} \right)} \right) =   \cr 
  &  = \left( {n + 3/2} \right)\left( {n - 1 + 3/2} \right) \cr} 
$$
so that the sum becomes
$$
\eqalign{
  & \sum\limits_{n = 0}^{49} {\left( {2n + 1} \right)\left( {2n + 3} \right)}  = 4\sum\limits_{n = 0}^{49} {\left( {n + 3/2} \right)^{\,\underline {\,2\,} } }  =   \cr 
  &  = {4 \over 3}\sum\limits_{n = 0}^{49} {\left( {\left( {n + 1 + 3/2} \right)^{\,\underline {\,3\,} }  - \left( {n + 3/2} \right)^{\,\underline {\,3\,} } } \right)}  =   \cr 
  &  = {4 \over 3}\left( {\left( {50 + 3/2} \right)^{\,\underline {\,3\,} }  - \left( {3/2} \right)^{\,\underline {\,3\,} } } \right) =   \cr 
  &  = {4 \over 3}\left( {\left( {50 + 3/2} \right)\left( {49 + 3/2} \right)\left( {48 + 3/2} \right) - \left( {3/2} \right)\left( {1/2} \right)\left( { - 1/2} \right)} \right) =   \cr 
  &  = {4 \over {3 \cdot 2^3 }}\left( {103 \cdot 101 \cdot 99 + 3} \right) = {{1029900} \over 6} = 171650 \cr} 
$$
