If the sum -K-1-1-K-2-K-K-K-2-a-b-c where a- b- c - N and lie in -1- 16- then a - b - c equals to

The correct option is D 11 Let T_k=(K+2)√(k) K√(K+2)/K(K+2)^2 K^2(K+2) =(K+2)√(K) K√(K+2)/2K(K+2)=1/2 [ 1/√(K) 1/√(K+2) ] ∴ T_1=1/2 [ 1/√(1 ...

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Standard XII

Mathematics

Higher Order Derivatives

If the sum ...

Question

If the sum $\infty \sum K = 1 \frac{1}{(K + 2) \sqrt{K} + K \sqrt{K + 2}} = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{c}}$ where $a, b, c \in N$ and lie in $[1, 16]$ , then $a + b + c$ equals to

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Solution

The correct option is D $11$
Let $T_{k} = \frac{(K + 2) \sqrt{k} - K \sqrt{K + 2}}{K (K + 2)^{2} - K^{2} (K + 2)}$

$= \frac{(K + 2) \sqrt{K} - K \sqrt{K + 2}}{2 K (K + 2)} = \frac{1}{2} [\frac{1}{\sqrt{K}} - \frac{1}{\sqrt{K + 2}}]$
$∴ T_{1} = \frac{1}{2} [\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{3}}]$
$T_{2} = \frac{1}{2} [\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{4}}]$
$T_{3} = \frac{1}{2} [\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{5}}]$ and so on
As K $\to \infty$ , sum $= \frac{1}{2} [1 + \frac{1}{\sqrt{2}}] = \frac{1 + \sqrt{2}}{2 \sqrt{2}}$
$= \frac{\sqrt{1} + \sqrt{2}}{\sqrt{8}}$
$\Rightarrow a + b + c = 11.$
Hence, option 'D' is correct.

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Similar questions

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If the sum ∞∑K=11(K+2)√K+K√K+2=√a+√b√c where a,b,c∈N and lie in [1,16], then a+b+c equals to

The correct option is D 11Let Tk=(K+2)√k−K√K+2K(K+2)2−K2(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] and so onAs K→∞, sum =12[1+1√2]=1+√22√2=√1+√2√8⇒a+b+c=11.Hence, option 'D' is correct.

If the sum $\infty \sum K = 1 \frac{1}{(K + 2) \sqrt{K} + K \sqrt{K + 2}} = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{c}}$ where $a, b, c \in N$ and lie in $[1, 16]$ , then $a + b + c$ equals to