It ain't about googling, it's about sharing and discussing; I'm sure that's why tuanl posts his problems here.
Use indicator variables: For each of the first \(n-1\) tosses, out of \(n\), let \(X_1=1\) if that toss and the next are TH, 0 otherwise. Then the expected number of TH's is equal to \(E(X_1+\cdots+X_{n-1})=(n-1)E(X_1)=(n-1)P(X_1=1)=\frac{n-1}{4}\). The same argument shows that the number of TT's is expected to be the same. So it's the variance that will decide.
Now for the variance. \(Var(X_1+\cdots+X_{n-1})=E((X_1+\cdots+X_{n-1})^2)-\large(\frac{n-1}{4}\right)^2\). \(E((X_1+\cdots+X_{n-1})^2)=(n-1)E(X_1^2)+2\sum_{i<j}E(X_iX_j)\). Clearly \((n-1)E(X_1^2)\) will be the same in both the TT and the TH cases. We need to see how the \(\sum E(X_iX_j)\) part differs.
1) TH case. When \(j-1=1\), \(X_iX_j\) clearly must equal 0. In all other cases, disjoint pairs of tosses are involved and these are independent, so \(E(X_iX_j)=P(X_iX_j=1)=P(X_i=1,X_j=1)=P(X_i=1)P(X_j=1)=\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}\).
2) TT case. This is clearly larger as the disjoint pairs of tosses give the same \(E(X_iX_j)\) as in the TT case, and the overlapping pairs give a positive \(E(X_iX_j)\), whereas in the TH case these are 0.
So it's better to pick TH.