Answer: (a) \(\binom{59}{r}2^{59-r}\). (b) \(r=19\).

(a) In the expansion of \((2+x)^{59}\), the general term containing \(x^r\) is

\(\binom{59}{r}2^{59-r}x^r.\)

Therefore the coefficient of \(x^r\) is

\(\binom{59}{r}2^{59-r}.\)

(b) The coefficient of \(x^{r+1}\) is

\(\binom{59}{r+1}2^{59-(r+1)}=\binom{59}{r+1}2^{58-r}.\)

Set the two coefficients equal:

\(\binom{59}{r}2^{59-r}=\binom{59}{r+1}2^{58-r}.\)

Divide both sides by \(\binom{59}{r}2^{58-r}\):

\(2=\dfrac{\binom{59}{r+1}}{\binom{59}{r}}.\)

Now

\(\dfrac{\binom{59}{r+1}}{\binom{59}{r}}=\dfrac{59-r}{r+1}.\)

So

\(2=\dfrac{59-r}{r+1}.\)

Multiplying by \(r+1\),

\(2r+2=59-r.\)

Thus

\(3r=57,\)

and hence

\(r=19.\)