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

Given ; nbsp;(2x^2+3x+4)^10 = nbsp; ∑_r=0^20 a_r x^r nbsp; …(1) Replace x by 2x in above identity: 2^10 (2x^2+3x+4)^10x^20 = ∑ _r=0^20a_r2^rx^r ⇒ 2^10 n ...

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Byju's Answer

Standard XII

Mathematics

Application of Multinomial Expansion

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

Suggest Corrections

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. If

P (x) = 2 x^{2} - 3 x + 4,

, then

P (- 1)

is equal to?

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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 identity: 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 & r=13 in R.H.S), we get a7a13=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 $a_{1} + a_{6} + a_{11} + a_{16} =$