Answer The Following Question ?
#1/(1 . 4)# +#1/(4 . 7)# +#1/(7 . 10)# #.........# Summation Of #n^(th)# term ??
2 Answers
Explanation:
The general term can be written:
#a_n = 1/((3n-2)(3n+1)) = 1/3(1/(3n-2)-1/(3n+1))#
So:
#sum_(n=1)^N a_n = sum_(n=1)^N 1/3(1/(3n-2)-1/(3n+1))#
#color(white)(sum_(n=1)^N a_n) = 1/3(sum_(n=1)^N 1/(3n-2)-sum_(n=1)^N 1/(3n+1))#
#color(white)(sum_(n=1)^N a_n) = 1/3(sum_(n=1)^N 1/(3n-2)-sum_(n=2)^(N+1) 1/(3n-2))#
#color(white)(sum_(n=1)^N a_n) = 1/3(1+color(red)(cancel(color(black)(sum_(n=2)^N 1/(3n-2))))-color(red)(cancel(color(black)(sum_(n=2)^N 1/(3n-2)))) - 1/(3N+1))#
#color(white)(sum_(n=1)^N a_n) = 1/3(1-1/(3N+1))#
#color(white)(sum_(n=1)^N a_n) = N/(3N+1)#
Explanation:
Let ,