Question

plz solve this math question​

Answer 1

Answer:-

$\Large\boxed{\quad\sf{(a)\quad\!10\ and\ 0}\quad}$

Solution:-

Given,

$\displaystyle\longrightarrow\sf{\sum_{k=4}^{143}\dfrac{1}{\sqrt k+\sqrt{k+1}}=a-\sqrt b\quad\quad\dots(1)}$

Consider the LHS,

$\displaystyle\longrightarrow\sf{\sum_{k=4}^{143}\dfrac{1}{\sqrt k+\sqrt{k+1}}}$

For rationalization, we have to multiply $\sf{\sqrt {k+1}-\sqrt k}$ to both the numerator and the denominator as follows,

$\displaystyle\longrightarrow\sf{\sum_{k=4}^{143}\dfrac{\sqrt{k+1}-\sqrt k}{\left(\sqrt {k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt k\right)}}$

We know the identity $\sf{(a+b)(a-b)=a^2-b^2.}$ So the denominator becomes,

$\displaystyle\longrightarrow\sf{\sum_{k=4}^{143}\dfrac{\sqrt{k+1}-\sqrt k}{\left(\sqrt{k+1}\right)^2-\big(\sqrt{k}\big)^2}}$

$\displaystyle\longrightarrow\sf{\sum_{k=4}^{143}\dfrac{\sqrt{k+1}-\sqrt k}{k+1-k}}$

$\displaystyle\longrightarrow\sf{\sum_{k=4}^{143}\left(\sqrt{k+1}-\sqrt k\right)}$

On expanding the sum, we get,

$\displaystyle\longrightarrow\sf{\!\sqrt{4+1}-\sqrt4+\sqrt{5+1}-\sqrt5+\!\dots\ \!\!\!+\sqrt{142+1}-\sqrt{142}+\sqrt{143+1}-\sqrt{143}}$

$\displaystyle\longrightarrow\sf{\sqrt5-\sqrt4+\sqrt6-\sqrt5+\!\dots\!+\sqrt{143}-\sqrt{142}+\sqrt{144}-\sqrt{143}}$

Corresponding additive inverse pairs are cancelled, then,

$\displaystyle\longrightarrow\sf{\sqrt{144}-\sqrt4}$

$\displaystyle\longrightarrow\sf{12-2}$

$\displaystyle\longrightarrow\sf{10}$

Hence (1) becomes,

$\displaystyle\longrightarrow\sf{10=a-\sqrt b}$

$\displaystyle\longrightarrow\sf{10-\sqrt0=a-\sqrt b}$

Therefore,

$\displaystyle\longrightarrow\sf{\underline{\underline{a=10}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{b=0}}}$

Hence (a) is the answer.

Answer 2

Answer:

(a) 10 and 0

Step-by-step explanation:

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

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

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

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

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

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

$\gray{\mathrm{Simplify}}$

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

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

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

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

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

$\dots$

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

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

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

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

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

$\mathrm{Simplify}$

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

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

$10 = a -\sqrt{b}$

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

$a=10,b=0$

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