Question

Solve the following equation:
$(x^2+x+1)+(x^2+2x+3)+(x^2+3x+5)+\cdots \cdots +(x^2+20x+39)=4500.$

Answer 1

The given equation is in the form of sum .

Total sum of $x^2$ is $20x^2$

Sum of $xto20x$ is $210x$ (by $\frac{n(n+1)}{2}$ formula)

Sum of $(1+3+\mathrm{5........39})$ find by sum of ap

Now, we have $n=20,a=1,d=2$

$s_n=\frac{n}{2}(2a+(n-1\left)d\right)$

=>$s_n=\frac{19}{2}(2\times 1+(20-1\left)2\right)$

$s_n=380$

now,

$20x^2+210x+380=4500$

$20x^2+210x-4120=0$

$=10.032,-20.5327$