<p><strong>If for some positive integer <math><mi>n</mi></math> , the coefficients of three consecutive terms in the binomial expansion of <math><mo>(</mo><mn>1</mn><mo>+</mo><mi>x</mi><msup><mrow><mrow><mo>)</mo></mrow></mrow><mrow><mrow><mi mathvariant="normal">N</mi><mo>+</mo><mn>5</mn></mrow></mrow></msup></math> are in the ratio <math><mn>5</mn><mo>:</mo><mn>10</mn><mo>:</mo><mn>14</mn></math> , then the largest coefficient in the expansion is:</strong></p>

f (2)=8, f' (2)=5, f' (x)≥1, f'' (x)≥4, ∀x∈ (1,6)Using LMVTf'' (x)=f' (5)-f' (2)5-2≥4⇒f' (5)≥17f' (x)=f (5)-f (2)5-2≥1⇒f (5)≥11Therefore f' (5)+f (5)≥28

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

If for some positive integer $n$ , the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{N + 5}$ are in the ratio $5 : 10 : 14$ , then the largest coefficient in the expansion is:

Option 1 -

252

Option 2 -

330

Option 3 -

792

Option 4 -

462(Binomial)

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Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

$f (2) = 8, f^{'} (2) = 5, f^{'} (x) \geq 1, f^{''} (x) \geq 4, \forall x \in (1,6)$

Using LMVT

$f^{''} (x) = \frac{f^{'} (5) - f^{'} (2)}{5 - 2} \geq 4 \Rightarrow f^{'} (5) \geq 17$

$f^{'} (x) = \frac{f (5) - f (2)}{5 - 2} \geq 1 \Rightarrow f (5) \geq 11$

Therefore $f^{'} (5) + f (5) \geq 28$

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

$\sum_{k = 1}^{10} \frac{k}{k^{4} + k^{2} + 1} = \frac{m}{n};$ 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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If for some positive integer $n$ , the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{N + 5}$ are in the ratio $5 : 10 : 14$ , then the largest coefficient in the expansion is:

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If for some positive integer n , the coefficients of three consecutive terms in the binomial expansion of (1+x)N+5 are in the ratio 5:10:14 , then the largest coefficient in the expansion is:

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If for some positive integer $n$ , the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{N + 5}$ are in the ratio $5 : 10 : 14$ , then the largest coefficient in the expansion is: