For a positive integer n, the \(n^{th}\) triangular number is \(T(n)=\dfrac{n(n+1)}{2}.\) For example, \(T(3) = \frac{3(3+1)}{2}= \frac{3(4)}{2}=6\), so the third triangular number is 6.
Determine the smallest integer \(b>2011\) such that \(T(b+1)-T(b)=T(x)\)for some positive integer x.
+26367 For a positive integer \(n\), the n^{th} triangular number is \(T(n)=\dfrac{n(n+1)}{2}\).
For example, \(T(3) = \frac{3(3+1)}{2}= \frac{3(4)}{2}=6\), so the third triangular number is \(6\).
Determine the smallest integer \(b>2011\) such that \(T(b+1)-T(b)=T(x)\) for some positive integer \(x\).
My attempt:
\(\begin{array}{|rcll|} \hline \mathbf{T(b+1)-T(b)} &=& \mathbf{T(x)} \\\\ \dfrac{(b+1)(b+1+1)}{2}-\dfrac{b(b+1)}{2} &=& \dfrac{x(x+1)}{2} \\\\ (b+1)(b+2)-b(b+1) &=& x(x+1) \\ (b+1)(b+2-b) &=& x(x+1) \\ 2(b+1) &=& x(x+1) \\ b+1 &=& \dfrac{x(x+1)}{2} \\ b+1 &=& T(x) \\ \mathbf{b} &=& \mathbf{T(x)-1} \qquad (1) \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline (1) & \mathbf{T(x)-1} &=& \mathbf{b} \quad | \quad b > 2011 \\ & T(x)-1 &\mathbf{>}& 2011 \\ & T(x) &\mathbf{>}& 2012 \\ & \dfrac{x(x+1)}{2}&\mathbf{>}& 2012 \\ & x(x+1) &\mathbf{>}& 4024 \\ & x^2+x &\mathbf{>}& 4024 \\ & x^2+x-4024 &\mathbf{>}& 0 \\\\ & \mathbf{x^2+x-4024} &=& \mathbf{0} \\\\ & x &=& \dfrac{-1+\sqrt{1-4(-4024)} }{2} \\ & x &=& \dfrac{-1+\sqrt{16097} }{2} \\ & x &=& \dfrac{-1+126.873953198}{2} \\ & x &=& \dfrac{125.873953198}{2} \\ & x &=& 62.9369765988 \\\\ & x &\mathbf{>}& 62.9369765988 \quad (\text{positive integer }x ) \\ & \mathbf{x} &=& \mathbf{63} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline T(63) &=& \dfrac{63(63+1)}{2} \\\\ T(63) &=& \dfrac{63*64}{2} \\\\ \mathbf{T(63)} &=& \mathbf{2016} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{b} &=& \mathbf{T(x) - 1} \quad | \quad x = 63 \\ b &=& T(63) - 1 \\ b &=& 2016 - 1 \\ \mathbf{b} &=& \mathbf{2015} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{T(b+1)} &=& \mathbf{ \dfrac{(b+1)(b+2)}{2} } \quad | \quad b = 2015 \\\\ T(2015+1) &=& \dfrac{(2015+1)(2015+2)}{2} \\\\ T(2016) &=& \dfrac{2016*2017}{2} \\\\ \mathbf{T(2016)} &=& \mathbf{2033136} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{T(b)} &=& \mathbf{ \dfrac{b(b+1)}{2} } \quad | \quad b = 2015 \\\\ T(2015) &=& \dfrac{2015(2015+1)}{2} \\\\ T(2016) &=& \dfrac{2015*2016}{2} \\\\ \mathbf{T(2015)} &=& \mathbf{2031120} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{T(b+1)-T(b)} &=& \mathbf{T(x)} \qquad b > 2011 \\ T(b+1)-T(b) &=& T(x) \quad | \quad b=2015,\ x = 63 \\ T(2016)-T(2015) &=& T(63) \quad | \quad \mathbf{T(2016)=2033136},\ \mathbf{T(2015)=2031120} \\ 2033136-2031120 &=& 2016\ \checkmark \\ \hline \end{array}\)
