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

Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial

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Vishal Baghel | Contributor-Level 10

2 months ago

$T_{5} =^{n} C_{4} {(2^{\frac{1}{4}})}^{n - 4} . {({(\frac{1}{3})}^{\frac{1}{4}})}^{4} =^{n} C_{4} . 2^{\frac{n - 4}{4}} . \frac{1}{3}$

$T_{6} =^{9} C_{5} {(2^{\frac{1}{2}})}^{4} . {({(\frac{1}{3})}^{\frac{1}{4}})}^{5} =^{9} C_{5} . 2 . \frac{1}{3^{5 / 4}} = \frac{2}{3} . \frac{1}{3^{\frac{1}{4}}} . \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$

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Similar Questions for you

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

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

alok kumar singh

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

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

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

$∴ m + n = 166$

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

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

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

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