If x2-x-1-x2-2x-3-x2-3x-5-x2-20x-39-4500 for x-0- then the value of x is

Given : (x^2+x+1)+(x^2+2x+3)+(x^2+3x+5)+⋯+(x^2+20x+39)=4500 ⇒ 20x^2+x(1+2+3+⋯+20)+(1+3+⋯+39)=4500 ⇒ 20x^2+x∑_n=1^20n+∑_n=1^20(2n 1)=4500 ⇒ 20x^2+x (20·212)+2( ...

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Byju's Answer

Standard XII

Mathematics

Sigma n

If x2+x+1+x2+...

Question

If $(x^{2} + x + 1) + (x^{2} + 2 x + 3) + (x^{2} + 3 x + 5) + \dots + (x^{2} + 20 x + 39) = 4500$ for $x > 0,$ then the value of $x$ is

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Solution

The correct option is B $10$
Given : $(x^{2} + x + 1) + (x^{2} + 2 x + 3) + (x^{2} + 3 x + 5) + \dots + (x^{2} + 20 x + 39) = 4500$
$\Rightarrow 20 x^{2} + x (1 + 2 + 3 + \dots + 20) + (1 + 3 + \dots + 39) = 4500$

$\Rightarrow 20 x^{2} + x 20 \sum n = 1 n + 20 \sum n = 1 (2 n - 1) = 4500$

$\Rightarrow 20 x^{2} + x (\frac{20 \cdot 21}{2}) + 2 (\frac{20 \cdot 21}{2}) - 20 = 4500$

$\Rightarrow 20 x^{2} + x (10 \cdot 21) + 20^{2} = 4500$
$\Rightarrow 2 x^{2} + 21 x - 450 + 40 = 0$
$\Rightarrow 2 x^{2} + 21 x - 410 = 0$
$\Rightarrow (x - 10) (2 x + 41) = 0$
$\Rightarrow x = 10$ or $x = - 20.5$
Since $x > 0,$ $∴ x = 10$

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If (x2+x+1)+(x2+2x+3)+(x2+3x+5)+⋯+(x2+20x+39)=4500 for x>0, then the value of x is

If $(x^{2} + x + 1) + (x^{2} + 2 x + 3) + (x^{2} + 3 x + 5) + \dots + (x^{2} + 20 x + 39) = 4500$ for $x > 0,$ then the value of $x$ is