What is expand n((ln(n+1))/(ln(n))-1) ?

The solution to expand n((ln(n+1))/(ln(n))-1) is (nln(n+1))/(ln(n))-n

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expand n((ln(n+1))/(ln(n))-1)

expand

n (\frac{ln ( n + 1 )}{ln ( n )} - 1)

\frac{n ln ( n + 1 )}{ln ( n )} - n

Solution steps

n (\frac{ln ( n + 1 )}{ln ( n )} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = n, b = \frac{ln ( n + 1 )}{ln ( n )}, c = 1

= n \frac{ln ( n + 1 )}{ln ( n )} - n \cdot 1

= n \frac{ln ( n + 1 )}{ln ( n )} - 1 \cdot n

Simplify

n \frac{ln ( n + 1 )}{ln ( n )} - 1 \cdot n : \frac{n ln ( n + 1 )}{ln ( n )} - n

= \frac{n ln ( n + 1 )}{ln ( n )} - n

\frac{( 2 x + 5 ) ^{2}}{4}

4!

6!

(4 + \frac{1}{8} x)^{6}

(3 - 1) (1 + 2)

What is expand n((ln(n+1))/(ln(n))-1) ?
The solution to expand n((ln(n+1))/(ln(n))-1) is (nln(n+1))/(ln(n))-n