The value of (1+2+3+·s+n) is equal to

Correct Answer is : log⁡n+log⁡(n+1)−log⁡2 n+(n+1)- 2logn+log(n+1)−log2

Correct Answer is : log⁡n+log⁡(n+1)−log⁡2 n+(n+1)- 2logn+log(n+1)−log2

Test Index

Logarithms

Question : 59 of 85

Marks: +1, -0

The value of

\log(1+2+3+\cdots+n)

is equal to

Solution:

Concept:

The sum of first

n

natural numbers is

\frac{n(n+1)}{2}

, and the logarithm of this product expands using properties of logarithms.

Explanation:

We know

1+2+3+\cdots+n = \frac{n(n+1)}{2}

\log(1+2+\cdots+n) = \log\left(\frac{n(n+1)}{2}\right)

Using

\log(ab) = \log a + \log b

and

\log(a/b) = \log a - \log b

, we get:

\log(n(n+1)) - \log 2 = \log n + \log(n+1) - \log 2

Answer:

Option B:

\log n + \log(n+1) - \log 2

Go to Question:

Prev Question

Next Question