Question

Sum of 5 terms ofthe progression log2+log4+log8+---


a)15log2
b)10log2
c)90log2
d)55log2​

Answer 1

Required answer

Choice (a) $15\log2$.

Solution

Let's find the sum of 5 terms.

Sum of five terms

= $\log2+\log4+\log8+\log16+\log32$

= $\log2^1+\log2^2+\log2^3+\log2^4+\log2^5$

= $1\log2+2\log2+3\log2+4\log2+5\log2$

= $(1+2+3+4+5)\log2$

= $\boxed{15\log2}$

Hence we showed that the sum of the progression is $15\log2$.

Answer 2

Answer:

Required Answer :-

Since we see that the number are multiplying by 2

$\bf log \: 2 + log \: 4 + log \: 8 + log \: 16 + \: log \: 32$

4 = 2²

8 = 2³

16 = 2⁴

32 = 2⁵

$\log \: {2}^{1} + \log {2}^{2} + \log \: {2}^{3} + \log \: {2}^{4} + \log {2}^{5}$

$1 \log \: 2 + 2 \log \: 2 + 3 \log \: 2 + 4 \log \: 2 + 5 \log2$

Taking log 2 as common

$(1 + 2 + 3 + 4 + 5) \log \: 2$

$(3 + 7 + 5) \log \: 2$

$15 \log \: 2$