For all positive integral values of n, the value of 3.1.2+3.2.3+3.3.4+...+3.n.(n+1) is
n(n+1)(n+2)
n(n+1)(2n+1)
(n-1)n(n+1)
(n-1)n(n+1)2
Explanation for correct option.
Step1. Finding nth term
Given, series 3.1.2+3.2.3+3.3.4+...+3.n.(n+1)
Therefore general term is Tr=3ĆrĆ(r+1)
Step2. Finding sum the series
Sum =ār=1nTr
S=ār=1nTr=3ār=1n(r2+r)=3{ār=1nr2+ār=1nr}=3{n(n+1)(2n+1)6+n(n+1)2}=3n(n+1){2n+16+12}=3n(n+1){2n+4}6=n(n+1)(n+2)
Hence, correct option is (A)
For all positive integral values of n, 32n - 2n + 1 is
divisible by
For every positive integral value of n, 3n > n3 when