You expect to receive $10,000 at graduation in two years. You plan on investing it at 11% until you have $75,000. How long will
1 answer:
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A)
To calculate this sum, we could use trigonometric identity:
We have:
![\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k+1-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{(k+1)^2-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\ ](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D-%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D%7D%7Bk%28k%2B1%29%7D-%5Cdfrac%7B%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%2B1-1%7D%7D%7Bk%28k%2B1%29%7D-%5Cdfrac%7B%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7B%28k%2B1%29%5E2-1%7D%7D%7Bk%28k%2B1%29%7D-%5Cdfrac%7B%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%0A)
![=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\dfrac{\sqrt{(k+1)^2-1}}{\sqrt{(k+1)^2}}-\dfrac{1}{k+1}\cdot\dfrac{\sqrt{k^2-1}}{\sqrt{k^2}}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{\dfrac{(k+1)&^2-1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{\dfrac{k^2-1}{k^2}}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\dfrac{1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{1-\dfrac{1}{k^2}}\right]=\\\\\\ ](https://tex.z-dn.net/?f=%3D%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B1%7D%7Bk%7D%5Ccdot%5Cdfrac%7B%5Csqrt%7B%28k%2B1%29%5E2-1%7D%7D%7B%5Csqrt%7B%28k%2B1%29%5E2%7D%7D-%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Ccdot%5Cdfrac%7B%5Csqrt%7Bk%5E2-1%7D%7D%7B%5Csqrt%7Bk%5E2%7D%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B1%7D%7Bk%7D%5Ccdot%5Csqrt%7B%5Cdfrac%7B%28k%2B1%29%26%5E2-1%7D%7B%28k%2B1%29%5E2%7D%7D-%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Ccdot%5Csqrt%7B%5Cdfrac%7Bk%5E2-1%7D%7Bk%5E2%7D%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B1%7D%7Bk%7D%5Ccdot%5Csqrt%7B1-%5Cdfrac%7B1%7D%7B%28k%2B1%29%5E2%7D%7D-%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Ccdot%5Csqrt%7B1-%5Cdfrac%7B1%7D%7Bk%5E2%7D%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%0A)
![=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\left(\dfrac{1}{k+1}\right)^2}-\dfrac{1}{k+1}\cdot\sqrt{1-\left(\dfrac{1}{k}\right)^2}\right]=\\\\\\= \sum\limits_{k=1}^n\left[\arcsin\left(\dfrac{1}{k}\right)-\arcsin\left(\dfrac{1}{k+1}\right)\right]=\\\\\\](https://tex.z-dn.net/?f=%3D%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B1%7D%7Bk%7D%5Ccdot%5Csqrt%7B1-%5Cleft%28%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Cright%29%5E2%7D-%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Ccdot%5Csqrt%7B1-%5Cleft%28%5Cdfrac%7B1%7D%7Bk%7D%5Cright%29%5E2%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Cleft%5B%5Carcsin%5Cleft%28%5Cdfrac%7B1%7D%7Bk%7D%5Cright%29-%5Carcsin%5Cleft%28%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Cright%29%5Cright%5D%3D%5C%5C%5C%5C%5C%5C)
![=\bigg[\arcsin(1)-\arcsin\left(\frac{1}{2}\right)\bigg]+\bigg[\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)\bigg]+\\\\\\+ \bigg[\arcsin\left(\frac{1}{3}\right)-\arcsin\left(\frac{1}{4}\right)\bigg]+\dots+ \bigg[\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)\bigg]=\\\\\\](https://tex.z-dn.net/?f=%3D%5Cbigg%5B%5Carcsin%281%29-%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cbigg%5D%2B%5Cbigg%5B%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29-%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7B3%7D%5Cright%29%5Cbigg%5D%2B%5C%5C%5C%5C%5C%5C%2B%0A%5Cbigg%5B%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7B3%7D%5Cright%29-%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7B4%7D%5Cright%29%5Cbigg%5D%2B%5Cdots%2B%0A%5Cbigg%5B%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7Bn%7D%5Cright%29-%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7Bn%2B1%7D%5Cright%29%5Cbigg%5D%3D%5C%5C%5C%5C%5C%5C)
So the answer is:
![\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)}](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D-%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Carcsin%5Cleft%28%5Cdfrac%7B1%7D%7Bn%2B1%7D%5Cright%29%7D)
B)
![\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \lim\limits_{n\to\infty}\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \lim\limits_{n\to\infty}\Bigg(\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)\Bigg)=\dfrac{\pi}{2}-\lim\limits_{n\to\infty}\arcsin\left(\dfrac{1}{n+1}\right)=\\\\\\= \Bigg\{\dfrac{1}{n+1}\xrightarrow{n\to\infty}0\Bigg\}=\dfrac{\pi}{2}-\arcsin(0)=\dfrac{\pi}{2}-0=\dfrac{\pi}{2}](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bk%3D1%7D%5E%5Cinfty%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D-%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7D%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D-%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7D%5CBigg%28%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Carcsin%5Cleft%28%5Cdfrac%7B1%7D%7Bn%2B1%7D%5Cright%29%5CBigg%29%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7D%5Carcsin%5Cleft%28%5Cdfrac%7B1%7D%7Bn%2B1%7D%5Cright%29%3D%5C%5C%5C%5C%5C%5C%3D%0A%5CBigg%5C%7B%5Cdfrac%7B1%7D%7Bn%2B1%7D%5Cxrightarrow%7Bn%5Cto%5Cinfty%7D0%5CBigg%5C%7D%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Carcsin%280%29%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D-0%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D)
So we prove that:
![\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bk%3D1%7D%5E%5Cinfty%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D-%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D)
To calculate this sum, we could use trigonometric identity:
We have:
So the answer is:
B)
So we prove that: