If -k - 1-1k - 2-k - k -k - 2 - -a - -b-c- where a- b- c - and a- b- c - -1- 15- then a - b -c is equal to

T_k = 1(k + 2)√(k) + k √(k + 2) = (k + 2)√(k) k √(k + 2)2k(k + 2) nbsp; = 12[1√(k) 1√(k+2)] T_1=12[1√(1) 1√(3)] T_2=12[1√(2) 1√(4)] T_3=12[1√(3) 1√ ...

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Standard XII

Mathematics

Infinite GP

If ∑k = 1∞1k ...

Question

If $\infty \sum k = 1 \frac{1}{(k + 2) \sqrt{k} + k \sqrt{k + 2}} = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{c}},$ where $a, b, c \in N$ and $a, b, c \in [1, 15],$ then $a + b + c$ is equal to

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Solution

$T_{k} = \frac{1}{(k + 2) \sqrt{k} + k \sqrt{k + 2}}$
$= \frac{(k + 2) \sqrt{k} - k \sqrt{k + 2}}{2 k (k + 2)}$
$= \frac{1}{2} [\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 2}}]$

$T_{1} = \frac{1}{2} [\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{3}}]$
$T_{2} = \frac{1}{2} [\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{4}}]$
$T_{3} = \frac{1}{2} [\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{5}}]$
$. . . . . .$

$∴$ Sum $= T_{1} + T_{2} + T_{3} \dots + T_{\infty}$
$= \frac{1}{2} [1 + \frac{1}{\sqrt{2}}] = \frac{\sqrt{2} + 1}{2 \sqrt{2}} = \frac{\sqrt{2} + \sqrt{1}}{\sqrt{8}}$
$\Rightarrow a + b + c = 11$

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If ∞∑k=11(k+2)√k+k√k+2=√a+√b√c, where a,b,c∈N and a,b,c∈[1,15], then a+b+c is equal to

Tk=1(k+2)√k+k√k+2 =(k+2)√k−k√k+22k(k+2) =12[1√k−1√k+2] T1=12[1√1−1√3] T2=12[1√2−1√4] T3=12[1√3−1√5] . .. .. . ∴ Sum =T1+T2+T3⋯+T∞ =12[1+1√2]=√2+12√2=√2+√1√8 ⇒a+b+c=11

If $\infty \sum k = 1 \frac{1}{(k + 2) \sqrt{k} + k \sqrt{k + 2}} = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{c}},$ where $a, b, c \in N$ and $a, b, c \in [1, 15],$ then $a + b + c$ is equal to