Building all three probabilities in terms of $k$ from the given recursive relation and using $\sum P(X=i)=1$ gives a quadratic in $k$, of which only the positive root is valid.

1. Given $P(X=0)=\dfrac17$.
2. For $i=1$: $P(X=1)=k\,P(X=0)=\dfrac k7$.
3. For $i=2$: $P(X=2)=k\,P(X=1)=k\cdot\dfrac k7=\dfrac{k^2}7$.
4. Since $X$ only takes values $0,1,2$, the probabilities must sum to $1$: $\dfrac17+\dfrac k7+\dfrac{k^2}7=1$. …

Building up $P(1)$ and $P(2)$ from $P(0)=1/7$ via the recurrence $P(i)=kP(i-1)$, then using $\sum P(x)=1$ to solve for $k$.

1. Given $P(X=0)=\dfrac17$ and $P(X=i)=kP(X=i-1)$ for $i=1,2$.
2. $P(X=1)=kP(X=0)=\dfrac{k}{7}$.
3. $P(X=2)=kP(X=1)=k\cdot\dfrac{k}{7}=\dfrac{k^2}{7}$. …

Using $\sum P(X=x)=1$ for the given pmf gives $\lambda=2$.

1. The probability mass function must satisfy $\displaystyle\sum_x P(X=x)=1$.
2. So $\dfrac{\lambda}{6}+\dfrac{\lambda}{4}+\dfrac{\lambda}{12}=1$.
3. Take LCM of $6,4,12$, which is $12$: $\dfrac{2\lambda}{12}+\dfrac{3\lambda}{12}+\dfrac{\lambda}{12}=1$.
4. Combine: $\dfrac{2\lambda+3\lambda+\lambda}{12}=\dfrac{6\lambda}{12}=1$. …

Using $\sum P(X=x)=1$ to find $a=\dfrac{1}{28}$, then $P(1\le X\le4)=14a=\dfrac12$.

1. Sum of all probabilities equals 1: $\dfrac14+2a+3a+4a+5a+\dfrac14=1$.
2. Simplify: $\dfrac12+14a=1\Rightarrow14a=\dfrac12\Rightarrow a=\dfrac{1}{28}$. …

Solving $\sum P(X=x)=1$ gives $k=\dfrac1{10}$ (the negative root is rejected as it makes some probabilities negative).

1. Sum all the probabilities and set the total equal to $1$: $0+k+2k+2k+3k+k^2+2k^2+(7k^2+k)=1$.
2. Collect linear terms: $k+2k+2k+3k+k=9k$. Collect quadratic terms: $k^2+2k^2+7k^2=10k^2$.
3. So $10k^2+9k=1$, i.e. $10k^2+9k-1=0$.
4. By the quadratic formula: $k=\dfrac{-9\pm\sqrt{81+40}}{20}=\dfrac{-9\pm\sqrt{121}}{20}=\dfrac{-9\pm11}{20}$.
5. This gives $k=\dfrac{2}{20}=\dfrac1{10}$ or $k=\dfrac{-20}{20}=-1$. …