What is the sum of series??

Question

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

Answer 1

Hint: Like Matti P. showed in the comments, the sum is $$\sum_{n=1}^{50}(2n+1)(2n-1)=\sum_{n=1}^{50}(4n^2-1)=4\sum_{n=1}^{50}n^2-50.$$ I'm sure you can take it from here.

Answer 2

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 $$ \eqalign{ & \sum\limits_{n = 0}^{49} {\left( {2n + 1} \right)\left( {2n + 3} \right)} = 4\sum\limits_{n = 0}^{49} {\left( {n + 3/2} \right)^{\,\underline {\,2\,} } } = \cr & = {4 \over 3}\sum\limits_{n = 0}^{49} {\left( {\left( {n + 1 + 3/2} \right)^{\,\underline {\,3\,} } - \left( {n + 3/2} \right)^{\,\underline {\,3\,} } } \right)} = \cr & = {4 \over 3}\left( {\left( {50 + 3/2} \right)^{\,\underline {\,3\,} } - \left( {3/2} \right)^{\,\underline {\,3\,} } } \right) = \cr & = {4 \over 3}\left( {\left( {50 + 3/2} \right)\left( {49 + 3/2} \right)\left( {48 + 3/2} \right) - \left( {3/2} \right)\left( {1/2} \right)\left( { - 1/2} \right)} \right) = \cr & = {4 \over {3 \cdot 2^3 }}\left( {103 \cdot 101 \cdot 99 + 3} \right) = {{1029900} \over 6} = 171650 \cr} $$