<math> <mrow> <mstyle displaystyle="true"> <munderover> <mo>∑</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </munderover> <mrow> <mfrac> <mrow> <mi>k</mi> </mrow> <mrow> <msup> <mrow> <mi>k</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mi>k</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>1</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>m</mi> </mrow> <mrow> <mi>n</mi> </mrow> </mfrac> <mo>;</mo> </mrow> </mstyle> </mrow> </math> 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

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Maths Class 12th Binomial Theorem Maths NCERT Exemplar Solutions Class 12th Chapter Eight

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

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alok kumar singh | Contributor-Level 10

2 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

$=^{15} C_{r} {(t^{2} x^{\frac{1}{5}})}^{15 - r} {(\frac{{(1 - x)}^{\frac{1}{10}}}{t})}^{r}$

for term independent on t

2 (15 – r) – r = 0

$\Rightarrow r = 10$

$∴ T_{11} =^{15} C_{10} \times (1 - x)$

Maximum value of x (1 – x) occur at

x = $\frac{1}{2}$

If the constant term in the expansion of ${(3 x^{3} - 2 x^{2} + \frac{5}{x^{5}})}^{10} i s 2^{k} . l,$ where $l$ is an odd integer, then the value of k is equal to:

alok kumar singh

${(3 x^{3} - 2 x^{2} + \frac{5}{x^{5}})}^{10}$

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

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

Now for constant term 3 + 2 $- 5 γ = 0$ …………..(i)

$α + β + γ = 10$ …………(ii)

From equation (i) & (ii)

$3 α + 2 (10 - α - γ) = 5 γ$

$α + 20 = 7 γ$

= 1, $γ$ = 3, = 6

Constant term $10! \cdot \frac{3^{1} \cdot {(- 2)}^{6} \cdot 5^{3}}{3! 6!} = 2^{9} \cdot 3^{2} \cdot 5^{4} \cdot 7^{1}$

Let the coefficients of the middle terms in the expansion of ${(\frac{1}{\sqrt[]{6}} + β x)}^{4}, {(1 - 3 β x)}^{2}$ and ${(1 - \frac{β}{2} x)}^{6}, β > 0,$ respectively form the first three terms of an A.P. If d is the common difference of this A.P., then 50 $- \frac{2 d}{β^{2}}$ is equal to………….

Vishal Baghel

Kindly consider the following figure

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

alok kumar singh

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

$\begin{array}{r} =^{50} C_{6} +^{49} C_{6} +^{48} C_{6} + \dots . + (^{30} C_{6} +^{30} C_{7}) -^{30} C_{7} \\ =^{50} C_{6} +^{49} C_{6} +^{48} C_{6} + \dots . + (^{31} C_{6} +^{31} C_{7}) -^{30} C_{7} \\ =^{50} C_{6} +^{50} C_{7} -^{30} C_{7} \\ =^{51} C_{7} -^{30} C_{7} \end{array}$

Let A₁, A₂, A₃,…. be an increasing geometric progression of positive real numbers. If A₁A₃A₅A₇ = $\frac{1}{1296}$ and A₂ + A₄ = $\frac{7}{36},$ then, the value of A₆ + A₈ + A₁₀ is equal to

Payal Gupta

Let $\frac{A_{2}}{A_{1}} = \frac{A_{3}}{A_{2}} = . . . = r$

$∴ A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$

$\Rightarrow A_{1} r^{3} = \frac{1}{6} . . . . . . . (i)$

Again, A₂ + A₄ = $\frac{7}{36}$

$A_{1} r = \frac{7}{36} - \frac{1}{6} = \frac{1}{36} . . . . . . . . (i i)$

$(i) & (i i) r = \sqrt[]{6} & A_{1} = \frac{1}{36 \sqrt[]{6}}$

$∴ A_{6} + A_{8} + A_{10} = A_{1} r^{5} (1 + r^{2} + r^{4}) = 43$

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