Question

plzzzz solve it..... ​

Answer 1

Answer:-

(a) 10 and 0

Step-by-step explanation:

$\displaystyle\sum _{k=4}^{143}\frac{1}{\sqrt{k}+\sqrt{k+1}}=a-\sqrt{b}$

$\mathrm{Let\:us\:consider\:the\:L.H.S;}$

$\displaystyle\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{k}-\sqrt{k+1}}{\sqrt{k}-\sqrt{k+1}}$

$= \displaystyle \sum^{143}_{k=4} \frac{1\cdot \left(\sqrt{k}-\sqrt{k+1}\right)}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}$

$= \displaystyle \sum^{143}_{k=4} \frac{\sqrt{k}-\sqrt{k+1}}{k-\left(k+1\right)}$

$= \displaystyle \sum^{143}_{k=4} \frac{\cancel{-}\left(\sqrt{k+1}-\sqrt{k}\right)}{\cancel{-}\left(\left(k+1\right)-k\right)}$

$\gray{\mathrm{Simplify}}$

$= \displaystyle \sum^{143}_{k=4} \frac{\sqrt{k+1}-\sqrt{k}}{1}$

$a_k=\sqrt{k+1}-\sqrt{k}$

$\mathrm{Compute\:a_4}:\quad \sqrt{5}-\sqrt{4}$

$\mathrm{Compute\:}a_5:\quad a_5=\sqrt{6}-\sqrt{5}$

$\mathrm{Compute\:}a_6:\quad a_6=\sqrt{7}-\sqrt{6}$

$\dots$

$\mathrm{Compute\:}a_{143}:\quad a_{143}=\sqrt{144}-\sqrt{143}$

$\mathrm{Expand\:the\:Sum}$

$\left(\sqrt{5}-\sqrt{4}\right)+\left(\sqrt{6}-\sqrt{5}\right)+\left(\sqrt{7}-\sqrt{6}\right)+\hdots+\left(\sqrt{144}-\sqrt{143}\right)$

$\implies \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\hdots+\sqrt{144}-\sqrt{143}$

$\implies -\sqrt{4}+\sqrt{144}$

$\mathrm{Simplify}$

$\implies -2+12 = \bold{10}$

$\mathrm{Compairing\:the\:L.H.S.\:with\:R.H.S};$

$10 = a -\sqrt{b}$

$\implies 10 - \sqrt{0} = a - \sqrt{b}$

$a=10,b=0$

$\mathrm{Option\:\left(a\right)~10~and~0}$

Answer 2

Answer:

Answer:-

(a) 10 and 0

Step-by-step explanation:

$\displaystyle\sum _{k=4}^{143}\frac{1}{\sqrt{k}+\sqrt{k+1}}=a-\sqrt{b}$

$\mathrm{Let\:us\:consider\:the\:L.H.S;}$

$\displaystyle\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{k}-\sqrt{k+1}}{\sqrt{k}-\sqrt{k+1}}$

$= \displaystyle \sum^{143}_{k=4} \frac{1\cdot \left(\sqrt{k}-\sqrt{k+1}\right)}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}$

$= \displaystyle \sum^{143}_{k=4} \frac{\sqrt{k}-\sqrt{k+1}}{k-\left(k+1\right)}$

$= \displaystyle \sum^{143}_{k=4} \frac{\cancel{-}\left(\sqrt{k+1}-\sqrt{k}\right)}{\cancel{-}\left(\left(k+1\right)-k\right)}$

$\blue{\mathrm{Simplify}}$

$= \displaystyle \sum^{143}_{k=4} \frac{\sqrt{k+1}-\sqrt{k}}{1}$

$a_k=\sqrt{k+1}-\sqrt{k}$

$\mathrm{Compute\:a_4}:\quad \sqrt{5}-\sqrt{4}$

$\mathrm{Compute\:}a_5:\quad a_5=\sqrt{6}-\sqrt{5}$

$\mathrm{Compute\:}a_6:\quad a_6=\sqrt{7}-\sqrt{6}$

$\dots$

$\mathrm{Compute\:}a_{143}:\quad a_{143}=\sqrt{144}-\sqrt{143}$

$\mathrm{Expand\:the\:Sum}$

$\left(\sqrt{5}-\sqrt{4}\right)+\left(\sqrt{6}-\sqrt{5}\right)+\left(\sqrt{7}-\sqrt{6}\right)+\left(\sqrt{144}-\sqrt{143}\right)$

$\implies \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\sqrt{144}-\sqrt{143}$

$\implies -\sqrt{4}+\sqrt{144}$

$\mathrm{Simplify}$

$\implies -2+12 = \bold{10}$

$\mathrm{Compairing\:the\:L.H.S.\:with\:R.H.S};$

$10 = a -\sqrt{b}$

$\implies 10 - \sqrt{0} = a - \sqrt{b}$

$a=10,b=0$

$\mathrm{Option\:\left(a\right)~10~and~0}$