∑ k = 1 1 0 k k 4 + k 2 + 1 = m n ; where m and n are co-prime, then m + n is equal to……..

∑k=110kk4+k2+1 =12∑k=110 [1k2−k+1−1k2+k+1] =12 [1−1111]=110222=55111=mn ∴m+n=166

$\sum_{k = 1}^{10} \frac{k}{k^{4} + k^{2} + 1} = \frac{m}{n};$ where m and n are co-prime, then m + n is equal to……..

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Alok Kumar Singh | Contributor-Level 10

7 months ago

$\sum_{k = 1}^{10} \frac{k}{k^{4} + k^{2} + 1}$

$= \frac{1}{2} \sum_{k = 1}^{10} [\frac{1}{k^{2} - k + 1} - \frac{1}{k^{2} + k + 1}]$

$= \frac{1}{2} [1 - \frac{1}{111}] = \frac{110}{222} = \frac{55}{111} = \frac{m}{n}$

$∴ m + n = 166$

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If the maximum value of the term independent of t in the expansion of ${(t^{2} x^{\frac{1}{5}} + \frac{{(1 - x)}^{\frac{1}{10}}}{t})}^{15}$ , x $\geq$ , is K, then 8 K is equal to…………..

Vishal Baghel

General term

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for term independent on t

2 (15 – r) – r = 0

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$= 10! \frac{3^{α} \cdot {(- 2)}^{β} \cdot 5^{γ} \cdot x^{3 α + 2 β - 5 γ}}{α! β! γ!}$

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$α + β + γ = 10$ …………(ii)

From equation (i) & (ii)

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Kindly consider the following figure

The value of $\sum_{r = 0}^{20}^{50 - r} C_{6}$ is equal to:

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$\sum_{r = 0}^{20} ?^{50 - r} C_{6} =^{50} C_{6} +^{49} C_{6} +^{48} C_{6} + \dots . +^{30} C_{6}$

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