<p><strong>Let <math><msup><mrow><mrow><mfenced separators="|"><mrow><mrow><mn>2</mn><msup><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><mn>3</mn><mi>x</mi><mo>+</mo><mn>4</mn></mrow></mrow></mfenced></mrow></mrow><mrow><mrow><mn>10</mn></mrow></mrow></msup><mo>=</mo><msubsup><mrow><mrow><mo>∑</mo></mrow></mrow><mrow><mrow><mi>r</mi><mo>=</mo><mn>0</mn></mrow></mrow><mrow><mrow><mn>20</mn></mrow></mrow></msubsup><mo>?</mo><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>r</mi></mrow></mrow></msub><msup><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mi>r</mi></mrow></mrow></msup></math> . Then <math><mfrac><mrow><mrow><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>7</mn></mrow></mrow></msub></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>13</mn></mrow></mrow></msub></mrow></mrow></mfrac></math> is equal to</strong></p>

Given 2x2+3x+410=∑r=020?arxrreplace x by 2x in above identity :-2102x2+3x+410x20=∑r=020??ar2rxr⇒210∑r=020??arxr=∑r=020??ar2rx(20-r)( from (i))now, comparing coefficient of x7 from both sides(take r=7 in L.H.S. and r=13 in R.H.S.)210a7=a13213⇒a7a13=23=8

Ask & Answer Question

Maths Class 12th Sequences and Series Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let ${(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} ? a_{r} x^{r}$ . Then $\frac{a_{7}}{a_{13}}$ is equal to

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Given ${(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} ? a_{r} x^{r}$

replace $x$ by $\frac{2}{x}$ in above identity :-

$\begin{array}{r} \frac{2^{10} {(2 x^{2} + 3 x + 4)}^{10}}{x^{20}} = \sum_{r = 0}^{20} ? ? \frac{a_{r} 2^{r}}{x^{r}} \\ \Rightarrow 2^{10} \sum_{r = 0}^{20} ? ? a_{r} x^{r} = \sum_{r = 0}^{20} ? ? a_{r} 2^{r} x^{(20 - r)} (from (i)) \end{array}$

now, comparing coefficient of $x^{7}$ from both sides

(take $r = 7$ in L.H.S. and $r = 13$ in R.H.S.)

$2^{10} a_{7} = a_{13} 2^{13} \Rightarrow \frac{a_{7}}{a_{13}} = 2^{3} = 8$

0 Upvotes 0 Downvotes

Similar Questions for you

If $1 + (1 - 2^{2} \cdot 1) + (1 - 4^{2} \cdot 3) + (1 - 6^{2} \cdot 5) + \dots \dots . + (1 - 20^{2} \cdot 19) = α - 220 β$ , then an ordered pair $(α, β)$ is equal to:

alok kumar singh

$1 + (1 - 2^{2} . 1) + (1 - 4^{2} . 3) + \dots \dots . + (1 - 20^{2} . 19)$

$= α - 220 β$

$= 11 - (2^{2} \cdot 1 + 4^{2} \cdot 3 + \dots . . + 20^{2} \cdot 19)$

$= 11 - 2^{2} \cdot \sum_{r = 1}^{10} ? r^{2} (2 r - 1) = 11 - 4 (\frac{110^{2}}{2} - 35 \times 11)$

$= 11 - 220 (103)$

$\Rightarrow α = 11, β = 103$

Let $a_{1}, a_{2}, \dots a_{n}$ be a given A.P. whose common difference is an integer and $S_{n} = a_{1} + a_{2}$ $+ \dots + a_{n}$ . If $a_{1} = 1, a_{1} = 300$ and $15 \leq n \leq 50$ , then the ordered pair $(S_{n - 4}, a_{n - 4})$ is equal to:

alok kumar singh

$x \frac{d y}{d x} - y = x^{2} (x c o s ? x + s i n ? x), x > 0$

$\frac{d y}{d x} - \frac{y}{x} = x (x c o s ? x + s i n ? x) \Rightarrow \frac{d y}{d x} + P y = Q$

So, I.F. $= e^{\int - \frac{1}{x} d x} \frac{1}{| x |} = \frac{1}{x} (x > 0)$

Thus, $\frac{y}{x} = \int \frac{1}{x} (x (x c o s ? x + s i n ? x)) d x$

$\Rightarrow \frac{y}{x} = x s i n ? x + C$

$? y (π) = π \Rightarrow C = 1$

So, $y = x^{2} s i n ? x + x \Rightarrow (y)_{π / 2} = \frac{π^{2}}{4} + \frac{π}{2}$

Also, $\frac{d y}{d x} = x^{2} c o s ? x + 2 x s i n ? x + 1$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - x^{2} s i n ? x + 4 x c o s ? x + 2 s i n ? x$

If a₁, a₂, a₃…. and b₁, b₂, b₃….. are A.P., and a₁ = 2, a₁₀ = 3, a₁b₁ = 1 – a₁₀b₁₀, then a₄b₄ is equal to……...

Payal Gupta

$a_{1}, a_{2}, a_{3} . . . . . .$ are in A.P. (Let common difference is d₁)

$b_{1}, b_{2}, b_{3} . . . .$ are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$? a_{1} b_{1} = 1$

$∴ b_{1} = \frac{1}{2}$

$a_{10} b_{10} = 1$

$∴ b_{10} = \frac{1}{3}$

Now,

$a_{10} = a_{1} + 9 d_{1} \Rightarrow d_{1} = \frac{1}{9}$

$b_{10} = b_{1} + 9 d_{2} \Rightarrow d_{2} = \frac{1}{9} [\frac{1}{3} - \frac{1}{2}] = - \frac{1}{54}$

Now

$a_{4} = 2 + \frac{3}{9} = \frac{7}{3}$

$b_{4} = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$

$∴ a_{4} b_{4} = \frac{28}{27}$

...more

$a_{1}, a_{2}, a_{3} . . . . . .$ are in A.P. (Let common difference is d₁)

$b_{1}, b_{2}, b_{3} . . . .$ are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$? a_{1} b_{1} = 1$

$∴ b_{1} = \frac{1}{2}$

$a_{10} b_{10} = 1$

$∴ b_{10} = \frac{1}{3}$

Now,

$a_{10} = a_{1} + 9 d_{1} \Rightarrow d_{1} = \frac{1}{9}$

$b_{10} = b_{1} + 9 d_{2} \Rightarrow d_{2} = \frac{1}{9} [\frac{1}{3} - \frac{1}{2}] = - \frac{1}{54}$

Now

$a_{4} = 2 + \frac{3}{9} = \frac{7}{3}$

$b_{4} = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$

$∴ a_{4} b_{4} = \frac{28}{27}$

less

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Let ${(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} ? a_{r} x^{r}$ . Then $\frac{a_{7}}{a_{13}}$ is equal to

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Let 2x2+3x+410=∑r=020?arxr . Then a7a13 is equal to

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Let ${(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} ? a_{r} x^{r}$ . Then $\frac{a_{7}}{a_{13}}$ is equal to