heureka

In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms.

The seventh term is equal to 6. Find the sum of all ten terms.

Let Fibonacci number sequence is:

\(\begin{array}{|rcll|} \hline \ldots \\ f_{-1} &=& 1 \\ f_{0} &=& 0 \\ f_{1} &=& 1 \\ f_{2} &=& 1 \\ f_{3} &=& 2 \\ f_{4} &=& 3 \\ f_{5} &=& 5 \\ f_{6} &=& 8 \\ f_{7} &=& 13 \\ f_{8} &=& 21 \\ f_{9} &=& 34 \\ f_{10} &=& 55 \\ f_{11} &=& 89 \\ f_{12} &=& 144 \\ \ldots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_n &=& f_{n-2}a_1+f_{n-1}a_2 \\ s_n &=& f_na_1+ \left(f_{n+1}-1 \right)a_2\\ \hline a_{7} = f_5a_1+f_6a_2 &=& 6 \\ f_6a_2 &=& 6-f_5a_1 \\ a_2 &=& \frac{6-f_5a_1}{f_6} \\\\ s_{10} &=& f_{10}a_1+ \left(f_{11}-1\right)a_2\\ &=& f_{10}a_1+ \left(f_{11}-1\right)\left(\frac{6-f_5a_1}{f_6}\right)\\ &=& f_{10}a_1+ \dfrac{6\left(f_{11}-1\right)}{f_6}- \dfrac{f_5\left(f_{11}-1\right)}{f_6}a_1 \\ &=& \left(f_{10}- \dfrac{f_5\left(f_{11}-1\right)}{f_6}\right)a_1+ \dfrac{6\left(f_{11}-1\right)}{f_6} \\ &=& \left(55- \dfrac{5\left(89-1\right)}{8}\right)a_1+ \dfrac{6\left(89-1\right)}{8} \\ &=& \left(55- \dfrac{5\cdot 88}{8}\right)a_1+ \dfrac{6\cdot 88}{8} \\ &=& (55- 55 )a_1+ 6\cdot 11 \\ &=& 0+66 \\ \mathbf{s_{10}} &\mathbf{=}& \mathbf{ 66 } \\ \hline \end{array}\)

help

\(\large{\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots}\)

\(\begin{array}{|rcll|} \hline && \dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots \\\\ &=& \dfrac{6}{(3-1)(3+1)}+\dfrac{6}{(5-1)(5+1)}+\dfrac{6}{(7-1)(7+1)}+\dfrac{6}{(9-1)(9+1)}+\cdots \\\\ &=& \dfrac{6}{2\cdot 4}+\dfrac{6}{4\cdot 6}+\dfrac{6}{6\cdot 8}+\dfrac{6}{8\cdot 10}+\cdots \\\\ &=& 6\left( \dfrac{1}{2\cdot 4}+\dfrac{1}{4\cdot 6}+\dfrac{1}{6\cdot 8}+\dfrac{1}{8\cdot 10}+\cdots \right) \\\\ &=& 6\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{2k(2k+2)} \\\\ &=& 6\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{4k(k+1)} \\\\ &=& \dfrac{6}{4}\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{k(k+1)} \\\\ &=& \dfrac{3}{2}\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{k(k+1)} \quad | \quad \boxed{\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1} } \\\\ &=& \dfrac{3}{2}\cdot \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k} - \dfrac{1}{k+1} \right) \\\\ &=& \dfrac{3}{2}\cdot \left[\sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k+1} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot \left[1+\sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k+1} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot \left[1+\sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot (1+0) \\\\ &=& \dfrac{3}{2} \\ \hline \end{array}\)

solutions

There are 21 possible sequences of steps that take the counter form square 1 to square 10.

\(\begin{array}{|r|l|} \hline 1.& (1\ 2\ 4\ 5\ 7\ 8\ 10) \\ 2.& (1\ 2\ 4\ 5\ 7\ 10) \\ 3.& (1\ 2\ 4\ 5\ 8\ 10) \\ 4.& (1\ 2\ 4\ 7\ 8\ 10) \\ 5.& (1\ 2\ 4\ 7\ 9\ 10) \\ 6.& (1\ 2\ 5\ 6\ 9\ 10) \\ 7.& (1\ 2\ 5\ 7\ 8\ 10) \\ 8.& (1\ 2\ 5\ 7\ 10) \\ 9.& (1\ 3\ 4\ 6\ 7\ 9\ 10) \\ 10.& (1\ 3\ 4\ 6\ 7\ 10) \\ 11.& (1\ 3\ 4\ 6\ 9\ 10) \\ 12.& (1\ 3\ 4\ 7\ 8\ 10) \\ 13.& (1\ 3\ 4\ 7\ 9\ 10) \\ 14.& (1\ 3\ 6\ 7\ 9\ 10) \\ 15.& (1\ 3\ 6\ 7\ 10) \\ 16.& (1\ 4\ 5\ 7\ 8\ 10) \\ 17.& (1\ 4\ 5\ 7\ 10) \\ 18.& (1\ 4\ 5\ 8\ 10) \\ 19.& (1\ 4\ 6\ 7\ 9\ 10) \\ 20.& (1\ 4\ 6\ 7\ 10) \\ 21.& (1\ 4\ 6\ 9\ 10) \\ \hline \end{array}\)

Math help

My attempt:

\(\begin{array}{|rcll|} \hline 2A &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \Big(1-2\sin(\phi)\Big)^2\ d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \left( 1-4\sin(\phi)+4\sin^2(\phi) \right) d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \left( 1\right)\ d\phi -\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } 4\sin(\phi)\ d\phi +\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } 4\sin^2(\phi) d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } d\phi -4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin(\phi)\ d\phi +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \left[\phi \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} -4\left[-\cos(\phi)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \left[\phi \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\left[\cos(\phi)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \dfrac{5\pi}{6}-\dfrac{\pi}{6} +4\left[ \underbrace{\cos\left(\frac{5\pi}{6}\right)-\cos\left(\frac{\pi}{6}\right) }_{=-\sqrt{3}} \right] +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \large{ \int \sin^2(\phi) } d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } \cos^2(\phi)\ d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } 1-\sin^2(\phi)\ d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } 1\ d\phi-\large{ \int } \sin^2(\phi)\ d\phi \\ 2\large{ \int \sin^2(\phi) } d\phi &=& -\cos(\phi)\sin(\phi)+ \phi \\ \mathbf{\large{ \int \sin^2(\phi) } d\phi} & \mathbf{=}& \mathbf{\dfrac12\Big(\phi-\cos(\phi)\sin(\phi) \Big) } \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline 2A &=& \dfrac{2\pi}{3} -4\sqrt{3} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ && \quad \mathbf{\large{ \int \sin^2(\phi) } d\phi=\dfrac12\Big(\phi-\cos(\phi)\sin(\phi) \Big) } \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \phi-\cos(\phi)\sin(\phi) \Big]_{\dfrac{\pi}{6}}^{\dfrac{5\pi}{6}} \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \phi-\dfrac12\sin(2\phi) \Big]_{\dfrac{\pi}{6}}^{\dfrac{5\pi}{6}} \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \dfrac{5\pi}{6}-\dfrac12\sin\left(\dfrac{10\pi}{6}\right)-\left(\dfrac{\pi}{6}-\dfrac12\sin\left(\dfrac{2\pi}{6}\right)\right) \Big] \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \dfrac{2\pi}{3}-\dfrac12\sin\left(\dfrac{5\pi}{3}\right)+\dfrac12\sin\left(\dfrac{ \pi}{3}\right) \Big] \\ &=& \dfrac{2\pi}{3} -4\sqrt{3}+2\left( \dfrac{2\pi}{3} + \dfrac{\sqrt{3}}{2} \right) \\ &=& \dfrac{6\pi}{3} -3\sqrt{3} \\ &=& 2\pi -3\sqrt{3} \\ &=& 1.08703288447 \\ A &=& \dfrac{1.08703288447 }{2} \\ \mathbf{A} & \mathbf{=} & \mathbf{0.54351644224} \\ \hline \end{array}\)

simplify

\(\mathbf{\large{\sqrt[3]{250x^6}}}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\large{\sqrt[3]{250x^6}}} \\ &=& \sqrt[3]{250}\sqrt[3]{x^6} \\ &=& \sqrt[3]{250}\cdot x^{\frac{6}{3}} \\ &=& \sqrt[3]{250}\cdot x^{2} \\ &=& \sqrt[3]{2\cdot125}\cdot x^{2} \\ &=& \sqrt[3]{2}\sqrt[3]{125}\cdot x^{2} \\ &=& \sqrt[3]{2}\sqrt[3]{5^3}\cdot x^{2} \\ &=& \sqrt[3]{2}\cdot 5^{\frac{3}{3}} \cdot x^{2} \\ &=& \sqrt[3]{2}\cdot 5^{1} \cdot x^{2} \\ &\mathbf{=}& \mathbf{\sqrt[3]{2}\cdot 5 \cdot x^{2} } \\ \hline \end{array}\)

Thank you, CPhill !

Two pedestrians standing 10 feet away and 40 feet away from the base of a billboard structure
have equivalent views of the billboard,
meaning that the angles marked in the diagram are congruent.
If the billboard is 9 feet tall, what is its height h off the ground?
hello! Can someone please solve this

\(\text{Let $\angle SPR = \angle SQR = \alpha $ } \\ \text{Let $\angle RPQ = \beta $ } \\ \text{Let $\angle RQB = \gamma $ } \)

\(\begin{array}{|lrcll|} \hline 1. & \tan(\beta) &=& \dfrac{h}{40} \\ & \beta &=& \arctan\left(\dfrac{h}{40}\right) \\ \hline 2. &\tan(\gamma) &=& \dfrac{h}{10} \\ & \gamma &=& \arctan\left(\dfrac{h}{10}\right) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline 3. & \tan(\alpha+\beta) &=& \dfrac{9+h}{40} \quad | \quad \beta = \arctan\left(\dfrac{h}{40}\right) \\ & \tan\left(\alpha+\arctan\left(\dfrac{h}{40}\right)\right) &=& \dfrac{9+h}{40} \\ & \alpha+\arctan\left(\dfrac{h}{40}\right) &=& \arctan\left(\dfrac{9+h}{40}\right) \\ & \alpha &=& \arctan\left(\dfrac{9+h}{40}\right)- \arctan\left(\dfrac{h}{40}\right) \\ \hline 4. & \tan(\alpha+\gamma)&=& \dfrac{9+h}{10} \quad | \quad \gamma = \arctan\left(\dfrac{h}{10}\right) \\ & \tan\left(\alpha+\arctan\left(\dfrac{h}{10}\right)\right)&=& \dfrac{9+h}{10} \\ & \alpha+\arctan\left(\dfrac{h}{10}\right)&=& \arctan\left(\dfrac{9+h}{10}\right) \\ & \alpha&=& \arctan\left(\dfrac{9+h}{10}\right)-\arctan\left(\dfrac{h}{10}\right) \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \text{Formula: } \arctan(x)-\arctan(y) = \arctan\left( \dfrac{x-y}{1+xy} \right) \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \alpha = \arctan\left(\dfrac{9+h}{40}\right)- \arctan\left(\dfrac{h}{40}\right) &=& \arctan\left(\dfrac{9+h}{10}\right)-\arctan\left(\dfrac{h}{10}\right) \\ \arctan\left(\dfrac{9+h}{40}\right)- \arctan\left(\dfrac{h}{40}\right) &=& \arctan\left(\dfrac{9+h}{10}\right)-\arctan\left(\dfrac{h}{10}\right) \\ \arctan\left( \dfrac{ \dfrac{(9+h)}{40}-\dfrac{h}{40} } { 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} } \right) &=& \arctan\left( \dfrac{ \dfrac{(9+h)}{10}-\dfrac{h}{10} } { 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} } \right) \\ \dfrac{ \dfrac{9}{40} } { 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} } &=& \dfrac{ \dfrac{9}{10} } { 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} } \\\\ \dfrac{9}{40} \left( 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} \right) &=& \dfrac{9}{10} \left( 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} \right) \\\\ 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} &=& 4 \left( 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} \right) \\ \dfrac{100+(9+h)h}{100} &=& 4\left( \dfrac{1600+(9+h)h}{1600} \right) \\ \dfrac{100+(9+h)h}{100} &=& \dfrac{1600+(9+h)h}{400} \\ 4\left( 100+(9+h)h \right) &=& 1600+(9+h)h \\ 400+4(9+h)h &=& 1600+(9+h)h \\ 400 +36h+4h^2 &=& 1600+9h+h^2 \\ 3h^2 +27h-1200 &=& 0 \quad | \quad : 3 \\ h^2 +9h -400 &=& 0 \\ h &=& \dfrac{-9\pm \sqrt{81-4\cdot(-400)} }{2} \\ h &=& \dfrac{-9\pm \sqrt{1681} }{2} \\ h &=& \dfrac{-9\pm 41 }{2} \\ h &=& \dfrac{-9{\color{red}\mathbf{+}} 41 }{2} \quad | \quad h>0!\\ \mathbf{h} &\mathbf{=}& \mathbf{16} \\ \hline \end{array}\)

Let t be a root of

\(f(x) = x^3 - x + 2\).

Evaluate

\(t^6 - t^2 + 4t\).

1.

\(\begin{array}{|rcll|} \hline f(t) = t^3 - t + 2 &=& 0 \\ t^3 - t + 2 &=& 0 \\ \mathbf{t^3} &\mathbf{=}&\mathbf{t-2} \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline t^6 - t^2 + 4t &=& (t^3)^2 - t^2 + 4t \quad | \quad \mathbf{t^3=t-2} \\ &=& (t-2)^2 - t^2 + 4t \\ &=& t^2-4t + 4 - t^2 + 4t \\ \mathbf{t^6 - t^2 + 4t} &\mathbf{=}&\mathbf{4} \\ \hline \end{array} \)

Let \(\mathbf{A}\) be a matrix, and let \(\mathbf{x}\) and \(\mathbf{y}\) be linearly independent vectors such that
\(\mathbf{A} \mathbf{x} = \mathbf{y}, \mathbf{A} \mathbf{y} = \mathbf{x} + 2\mathbf{y}\)
Then we have that
\(\mathbf{A}^{-1} \mathbf{x} = a \mathbf{x} + b\mathbf{y}\)
or some scalars a and b.

Find the ordered pair (a,b).

\(\begin{array}{|rcll|} \hline Ax &=& y \quad | \quad \cdot A^{-1} \\ A^{-1}Ax &=& A^{-1}y \quad | \quad A^{-1}A = I \\ x &=& A^{-1}y \\ \mathbf{A^{-1}y} &\mathbf{=}&\mathbf{x} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline Ay &=& x + 2y \quad | \quad \cdot A^{-1} \\ A^{-1}Ay &=& A^{-1}x + 2A^{-1}y \quad | \quad A^{-1}A = I \\ y &=& A^{-1}x + 2A^{-1}y \quad | \quad \mathbf{A^{-1}y=x} \\ y &=& A^{-1}x + 2x \\ \mathbf{A^{-1}x} &\mathbf{=}&\mathbf{-2x+y } \\ \hline \end{array}\)

\(\mathbf{(a,b) = (-2,1)}\)

Compute

\(\mathbf{\large{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}}}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{n^3-n} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{n(n^2-1)} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{n(n-1)(n+1)} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{(n-1)n(n+1)} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{n(n+1)(n+2)} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{n(n+1)}\cdot \dfrac{1}{(n+2)} \quad | \quad \boxed{\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}} \\\\ &=& \sum \limits_{n = 1}^{99} \left(\dfrac{1}{n} - \dfrac{1}{n+1} \right) \cdot \dfrac{1}{(n+2)} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{n(n+2)} - \dfrac{1}{(n+1)(n+2)} \quad | \quad \boxed{\dfrac{1}{n(n+2)} = \dfrac{1}{2}\left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) } \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2}\left(\dfrac{1}{n} - \dfrac{1}{n+2} \right) -\dfrac{1}{(n+1)(n+2)} \quad | \quad \boxed{\dfrac{1}{(n+1)(n+2)}= \dfrac{1}{n+1} - \dfrac{1}{n+2} } \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2}\left(\dfrac{1}{n} - \dfrac{1}{n+2} \right) -\left( \dfrac{1}{n+1} - \dfrac{1}{n+2} \right)\\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{2(n+2)} - \dfrac{1}{n+1} + \dfrac{1}{n+2}\\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{n+2} \left(1 - \dfrac{1}{2} \right) \\\\ &\mathbf{=}& \mathbf{\sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &\mathbf{=}& \mathbf{\sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)}} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \sum \limits_{n = 1}^{99} \dfrac{1}{n+1} + \sum \limits_{n = 1}^{99} \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{2}\sum \limits_{n = 1}^{99} \dfrac{1}{n} - \sum \limits_{n = 1}^{99} \dfrac{1}{n+1} + \dfrac{1}{2}\sum \limits_{n = 1}^{99} \dfrac{1}{n+2} \\\\ && \boxed{\dfrac{1}{2}\sum \limits_{n = 1}^{99} \dfrac{1}{n} \\ = \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 3}^{99} \dfrac{1}{n} \\ = \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} } \\ && \boxed{\sum \limits_{n = 1}^{99} \dfrac{1}{n+1} \\ = \dfrac12 + \sum \limits_{n = 2}^{99} \dfrac{1}{n+1} \\ = \dfrac12 + \sum \limits_{n = 1}^{98} \dfrac{1}{n+2} \\ = \dfrac12 + \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} + \dfrac{1}{100} } \\ && \boxed{ \dfrac{1}{2}\sum \limits_{n = 1}^{99} \dfrac{1}{n+2} \\ = \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} } \\\\ &=& \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\ && - \left( \dfrac12 + \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} + \dfrac{1}{100} \right) \\ && + \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} \\\\ &=& \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\ && - \dfrac12 - \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} - \dfrac{1}{100} \\ && + \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} \\\\ &=& \dfrac14- \dfrac{1}{100} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101}\\ && +\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} - \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} + \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\\\ &=& \dfrac14- \dfrac{1}{100} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101}+ \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} - \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\\\ &=& \dfrac14- \dfrac{1}{100} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} \\\\ &\mathbf{=}& \mathbf{\dfrac12 \left( \dfrac12- \dfrac{1}{50} +\dfrac{1}{100}+\dfrac{1}{101} \right)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &\mathbf{=}& \mathbf{\dfrac12 \left( \dfrac12- \dfrac{1}{50} +\dfrac{1}{100}+\dfrac{1}{101} \right)} \\\\ &\mathbf{=}& \dfrac{5049}{20200} \\\\ &\mathbf{=}& \mathbf{0.24995049505} \\ \hline \end{array}\)