Let , +2, Z, be the roots of the quadratic equation x(x+2)+(x+1)(x+3)+(x+2)(x+4)+ .+(x+n-1…

Question

Let $\alpha ,\alpha +2,\alpha \in Z$, be the roots of the quadratic equation $x(x+2)+(x+1)(x+3)+(x+2)(x+4)+\ldots .+(x+n-1)(x+n+1)=4n$ for some $n\in N$. Then $n+\alpha $ is equal to :

Accepted Answer

We are given that

the quadratic equation is

$(x)(x+2)+(x+1)(x+3)+(x+2)(x+4)+\ldots +(x+n-1)(x+n+1)=4n$

and the roots are $\alpha ,\alpha +2$ (both integers).

We need to find the value of $n+\alpha $.

Expand the general term: The ${k}^{th}$ term of the left-hand side is:

$(x+k)(x+k+2)$

Expanding this:

$(x+k)(x+k+2)={x}^{2}+2(k+1)x+k(k+2)$

Sum over all $k$ from $0$ to $n-1$: $=\sum _{n-1}^{k=0}[{x}^{2}+2(k+1)x+k(k+2)]$

Let's compute each part separately.

For ${x}^{2}$:

$\sum _{n-1}^{k=0}{x}^{2}=n{x}^{2}$

For $2(k+1)x$:

$\sum _{n-1}^{k=0}2(k+1)x=2x\sum _{n-1}^{k=0}(k+1)=2x\cdot \frac{n(n+1)}{2}=n(n+1)x$

For $k(k+2)$:

$\sum _{n-1}^{k=0}k(k+2)=\sum _{n-1}^{k=0}({k}^{2}+2k)=\sum {k}^{2}+2\sum k$

Now,

$\sum _{n-1}^{k=0}k=\frac{n(n-1)}{2},\, \sum _{n-1}^{k=0}{k}^{2}=\frac{(n-1)n(2n-1)}{6}$

Therefore,

$\sum _{n-1}^{k=0}k(k+2)=\frac{(n-1)n(2n-1)}{6}+n(n-1)$

Simplify:

$=n(n-1)(\frac{2n-1}{6}+1)=n(n-1)(\frac{2n-1+6}{6})=\frac{n(n-1)(2n+5)}{6}$

Combine all terms: $=n{x}^{2}+n(n+1)x+\frac{n(n-1)(2n+5)}{6}$

The equation is

$n{x}^{2}+n(n+1)x+\frac{n(n-1)(2n+5)}{6}=4n$

Simplify by dividing both sides by $n$ (since $n>0$): ${x}^{2}+(n+1)x+\frac{(n-1)(2n+5)}{6}=4$

${x}^{2}+(n+1)x+[\frac{(n-1)(2n+5)}{6}-4]=0$

Simplify the constant term:

$\frac{(n-1)(2n+5)}{6}-4=\frac{(n-1)(2n+5)-24}{6}=\frac{2{n}^{2}+3n-29}{6}$

Thus, the equation is

${x}^{2}+(n+1)x+\frac{2{n}^{2}+3n-29}{6}=0$

Use relation between roots and coefficients: Let roots be $\alpha $ and $\alpha +2$.

Then,

Sum of roots:

$\alpha +(\alpha +2)=2\alpha +2=-(n+1)$

so,

$\alpha =\frac{-(n+1)-2}{2}=-\frac{n+3}{2}$

Product of roots:

$\alpha (\alpha +2)=\frac{2{n}^{2}+3n-29}{6}$

Substitute $\alpha =-\frac{n+3}{2}$:

$\alpha (\alpha +2)=(-\frac{n+3}{2})(-\frac{n-1}{2})=\frac{(n+3)(n-1)}{4}$

Equate the two expressions for product: $\frac{(n+3)(n-1)}{4}=\frac{2{n}^{2}+3n-29}{6}$

Simplify:

$3(n+3)(n-1)=2(2{n}^{2}+3n-29)$

Expand:

$3({n}^{2}+2n-3)=4{n}^{2}+6n-58$

$3{n}^{2}+6n-9=4{n}^{2}+6n-58$

Simplify:

$3{n}^{2}+6n-9-4{n}^{2}-6n+58=0$

$-{n}^{2}+49=0\, \implies \, {n}^{2}=49$

$n=7n\in N)$

Find $\alpha $ and hence $n+\alpha $: $\alpha =-\frac{n+3}{2}=-\frac{7+3}{2}=-5$

$n+\alpha =7+(-5)=2$

$n+\alpha =2$

Correct Option: C (2)

Solutions · 1