Question

Prove that : log a + log a²+ log a³+ ... + log a^n = n(n+1)/2 log a

Answer 1

Step-by-step explanation:

To prove:

$log \: a + log \: {a}^{2} + log \: {a}^{3} + ... + log \: {a}^{n} = \frac{n(n + 1)}{2} log \: a$

Proof:

$log \: a + log \: {a}^{2} + log \: {a}^{3} + ... + log \: {a}^{n} \\ = log \: a \: + 2log \: a + 3log \: a + ... + nlog \: a \\ = log \: a(1 + 2 + 3 + ... + n) \\ = log \: a \times \frac{n( n+ 1)}{2} \\ = \frac{n(n + 1)}{2} \times log \: a$

Logarithm identities

$log_{b} \: a = \frac{1}{ log_{a} \: \: b } = \frac{ log_{m} \: a }{ log_{m} \: b } \\ log_{n}(ab) = l og_{n} \: a + log_{n} \: b \\ log_{n}( \frac{a}{b} ) = log_{n} \: a - log_{n} \: b \\ log_{a}( {b}^{n} ) = n \: log_{a} \: b \\ log_{a} \: a = 1$

Arithmetic Progression

a = first term

d = common difference

$a_{n } = {n}^{th} \: term \: \\ s_{n} = sum \: of \: n \: terms$

$a_{n} \: = a + (n - 1)d \\ s_{n} = \frac{n}{2} (2a + ( n- 1)d)$