Question

2 log a + 2 log a^2 + 2 log a^3 + ..... +2 log a^m​

Answer 1

Given,

• 2 log a + 2 log a² + 2 log a³ + ... + 2 log $\sf{a^m}$

We know,

• log $\sf{a^m}$ = m log a

Here,

• We can write log a² as 2 log a and so on.

Therefore,

• On taking log a common, we get

2 ( 1 + 2 + 3 + 4 + . . . . . + n ) log a

We know,

• Sum of 2n natural numbers is  n ( 2n + 1 )

Therefore,

$\boxed{\bold{2 \:log \:a + 2\: log\: a^2 + 2\: log\: a^3 + ... + 2\: log\: a^m = m\: ( 2m+ 1 )\: log \:a}}$