Question

loga + loga²+ loga³ + ... + loga n
= 1/2 n(n+1) loga
​

Answer 1

Step-by-step explanation:

loga + loga²+ loga³ + ... + loga n

= log a + 2 loga +3loga +......+ n log a ( were we know logX²= 2logx)

=log a(1+2+3+...+n)

=loga {n(n+1)2} [we know sum of frist n th natural number's is n(n+1)/2]

= 1/2 n(n+1) log a