Let 2x2-3x-410-r-020arxr- Then a7a13 is equal to

Given (2x^2+3x+4)^10=∑_r=0^20a_rx^r ⋯(1) Replace x by 2x in above, we get 2^10(2x^2+3x+4)^10x^20=∑_r=0^20a_r2^rx^r ⇒ 2^10∑_r=0^20a_rx^r=∑_r=0^20a_r2^rx^(20 r) ...

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Byju's Answer

Standard XII

Mathematics

Finding Remainder of Numbers of the Form a^b

Let 2x2+3x+41...

Question

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

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Solution

Given $(2 x^{2} + 3 x + 4)^{10} = 20 \sum r = 0 a_{r} x^{r} \dots (1)$
Replace $x$ by $\frac{2}{x}$ in above, we get
$\frac{2^{10} (2 x^{2} + 3 x + 4)^{10}}{x^{20}} = 20 \sum r = 0 \frac{a_{r} 2^{r}}{x^{r}}$
$\Rightarrow 2^{10} 20 \sum r = 0 a_{r} x^{r} = 20 \sum r = 0 a_{r} 2^{r} x^{(20 - r)}$ (from eq. $(1)$ )
Now, comparing coefficient of $x^{7}$ from both sides
(take $r = 7$ in L.H.S. and $r = 13$ in R.H.S.)
$2^{10} a_{7} = a_{13} 2^{13} \Rightarrow \frac{a_{7}}{a_{13}} = 2^{3} = 8$

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then $a_{1} + a_{6} + a_{11} + a_{16} =$

Q. Let

p, q, r

be roots of cubic equation

x^{3} + 2 x^{2} + 3 x + 3 = 0

, then

Q. Let

f : (0, \infty) \to R

be defined as

f (x) = {tan}^{- 1} ({log}_{2} (x))

and

g : R \to R

be defined as

g (x) = \frac{2 x^{2} - 3 x + 2}{2 x^{2} + 3 x + 2}

f \circ g

denotes the composition of

f

and

g,

then

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Let (2x2+3x+4)10=20∑r=0arxr. Then a7a13 is equal to

Given (2x2+3x+4)10=20∑r=0arxr ⋯(1) Replace x by 2x in above, we get 210(2x2+3x+4)10x20=20∑r=0ar2rxr ⇒21020∑r=0arxr=20∑r=0ar2rx(20−r) (from eq.(1)) 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

then a1+a6+a11+a16=

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

then $a_{1} + a_{6} + a_{11} + a_{16} =$