 K (John Abraham) is so addicted to smoking that is has become impossible for him to live without it. However, he gets a rude jolt when fed up with his craving.Q: Help in solving this family of equations Solve the family of equations: $\sum _{i=0}^{k}\frac{a^{i}\left( \frac{1}{k}% \sum\limits_{i=0}^{k-1}a^i\right)^{i+1}}{i+1}x^i=1$ where $a$ and $k$ are constants. thanks in advance. A: I think there are some errors in your initial formula, as normally there is no $k$ in the denominator. Anyway, here is how I would proceed: Let $F(a,x)=\sum_{i=0}^k \frac{a^i(1/k\sum_{i=0}^{k-1}a^i)^{i+1}}{i+1}x^i$. Then $$\frac{\partial F}{\partial x}(1,1)=\frac{1}{k}\sum_{i=0}^{k-1}a^i=1$$ and $$F(a,x)=1+\sum_{i=1}^k \frac{a^i(1/k\sum_{i=0}^{k-1}a^i)^{i+1}}{i+1}x^i$$ We notice that $$\sum_{i=0}^k \frac{a^{i+1}}{i+1}=\frac{1-a^k}{k(1-a)}$$ and hence we have $$F(a,x)=1+\frac{1-a^k}{k(1-a)}x+\sum_{i=2}^k \frac{a^i(1/k\sum_{i=0}^{k-1}a^i)^{i+1}}{i+1}x^i$$ We can now cancel the factor $1/k$ and take the $a$th root. We find F(a,x)=\frac{1-a^k