Binomial Theorem Question Pls Help

If \(n\) is a positive integer and \((x+1)^n\) is expanded in decreasing powers of \(x\),
three consecutive numerical coefficients are in the ratio \(2:15:70\). Compute \(n\).

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k + 1}} &=& \dfrac{n!}{(k+1)!(n-k-1)!} \quad &| \quad k!(k+1)=(k+1)! \\\\ &=& \dfrac{1}{k+1} *\dfrac{n!}{k!(n-k-1)!} \quad &| \quad (n-k-1)!(n-k)=(n-k)! \\\\ &=& \dfrac{n-k}{k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{n-k}{k+1} \dbinom{n}{k} } \\ \hline \end{array} \)

similarly \( \mathbf{\dbinom{n}{k - 1} = \dfrac{k}{n-k+1} \dbinom{n}{k} }\)

\(\begin{array}{|rcll|} \hline \dbinom{n}{k - 1} : \dbinom{n}{k} : \dbinom{n}{k + 1} &=& 2:15:70 \\\\ &&\boxed{ \mathbf{\dbinom{n}{k - 1} = \dfrac{k}{n-k+1} \dbinom{n}{k} }\\ ~\\ \mathbf{\dbinom{n}{k + 1} = \dfrac{n-k}{k+1} \dbinom{n}{k} } } \\\\ \dfrac{k}{n-k+1} \dbinom{n}{k} : \dbinom{n}{k} : \dfrac{n-k}{k+1} \dbinom{n}{k} &=& 2:15:70 \\ \hline \dfrac{\dbinom{n}{k}}{\dfrac{k}{n-k+1} \dbinom{n}{k}} &=& \dfrac{15}{2} \\\\ \dfrac{n-k+1}{k} &=& \dfrac{15}{2} \\\\ n-k+1 &=& \dfrac{15}{2}k \\ n &=& \dfrac{15}{2}k +k-1\\ \mathbf{n} &=& \mathbf{\dfrac{17}{2}k -1} \qquad (1) \\ \hline \dfrac{\dfrac{n-k}{k+1} \dbinom{n}{k} } {\dbinom{n}{k}} &=& \dfrac{70}{15} \\\\ \dfrac{n-k}{k+1} &=& \dfrac{70}{15} \\\\ 70(k+1) &=& 15(n-k) \\ 70k+70 &=& 15n-15k \\ 85k &=& 15n-70 \\ \mathbf{k} &=& \mathbf{ \dfrac{15n-70}{85} } \qquad (2) \\ \hline n &=& \dfrac{17}{2}k -1 \quad | \quad k=\dfrac{15n-70}{85} \\\\ n &=& \dfrac{17(15n-70)}{2*85}-1 \\\\ n &=& \dfrac{17(15n-70)}{170}-1 \\\\ n &=& \dfrac{ 15n-70 }{10}-1 \quad | \quad *10 \\\\ 10n &=& 15n-70-10 \\ 10n &=& 15n-80 \\ 5n &=& 80 \\ n &=& \dfrac{80}{5}\\ \mathbf{n} &=& \mathbf{16} \\\\ k &=& \dfrac{15n-70}{85} \quad | \quad n=16 \\\\ k &=& \dfrac{15*16-70}{85} \\ k &=& \dfrac{170}{85} \\ \mathbf{k} &=& \mathbf{2} \\ \hline \dbinom{n}{k - 1} : \dbinom{n}{k} : \dbinom{n}{k + 1} &=& 2:15:70 \\\\ \dbinom{16}{1} : \dbinom{16}{2} : \dbinom{16}{3} &=& 2:15:70 \\\\ 16 : 120 : 560 &=& 2:15:70 \\ \hline \end{array}\)