If `0 lt a,b lt 1 and tan^(-1)a+tan^(-1) b = pi/4`, then the value of
`(a+b)-((a^2+b^2)/2)+((a^3+b^3)/3)-((a^4+b^4)/4)+...` is :

Options

A. `log_e2`
B. `e^2-1`
C. e
D. `log_e(e/2)`

Answer

To solve the problem, we start with the given condition:

\[ \tan^{-1} a + \tan^{-1} b = \frac{\pi}{4} \]

Using the tangent addition formula, we know that:

\[ \tan(\tan^{-1} a + \tan^{-1} b) = \frac{a + b}{1 - ab} \]

Setting this equal to \(\tan\left(\frac{\pi}{4}\right) = 1\), we have:

\[ \frac{a + b}{1 - ab} = 1 \]

Cross-multiplying gives:

\[ a + b = 1 - ab \]

Rearranging this, we find:

\[ a + b + ab = 1 \]

Now, we need to evaluate the series:

\[ S = (a+b) - \frac{(a^2+b^2)}{2} + \frac{(a^3+b^3)}{3} - \frac{(a^4+b^4)}{4} + \ldots \]

We can express \(S\) in a more manageable form. Notice that:

\[ a^2 + b^2 = (a+b)^2 - 2ab \]

Substituting \(a+b = 1 - ab\):

\[ a^2 + b^2 = (1 - ab)^2 - 2ab = 1 - 2ab + a^2b^2 - 2ab = 1 - 4ab + a^2b^2 \]

Now, we can rewrite the series \(S\):

\[ S = (a+b) - \frac{(1 - 4ab + a^2b^2)}{2} + \frac{(a^3 + b^3)}{3} - \frac{(a^4 + b^4)}{4} + \ldots \]

Next, we can use the identity for the sum of cubes:

\[ a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (1 - ab)((1 - 4ab + a^2b^2) - ab) \]

Continuing this process, we can recognize that the series can be expressed in terms of logarithmic functions. Specifically, we can relate the series to:

\[ S = \ln(1 + a) + \ln(1 + b) = \ln((1 + a)(1 + b)) \]

Using the earlier result \(a + b + ab = 1\), we can substitute \(1 + a + b + ab = 2\):

\[ S = \ln(2) \]

Thus, the final value of the series is:

\[ \boxed{\ln(2)} \]