If, for a positive integer n, the quadratic equation x(x+1)+(x+1)(x+2)++(x+n-1)(x+n)=10n h…

Question

If, for a positive integer $n$, the quadratic equation $x(x+1)+(x+1)(x+2)+\cdots+(x+n-1)(x+n)=10n$ has two consecutive integral solutions, then $n$ is equal to:

Accepted Answer

$$\sum\limits_{r = 1}^n {\left( {x + r - 1} \right)\left( {x + r} \right)}  = 10n$$

$$ \Rightarrow $$ $$\sum\limits_{r = 1}^n {\left( {{x^2} + xr + \left( {r - 1} \right)x + {r^2} - r} \right)}  = 10n$$

$$ \Rightarrow $$ $$\sum\limits_{r = 1}^n {\left( {{x^2} + \left( {2r - 1} \right)x + r\left( {r - 1} \right)} \right)}  = 10n$$

$$ \Rightarrow $$ $$n{x^2} + \left\{ {1 + 3 + 5 + .... + \left( {2n - 1} \right)} \right\}x$$

$$ + \left\{ {1.2 + 2.3 + ... + \left( {n - 1} \right)n} \right\}$$ = 10n

$$ \Rightarrow $$ $$n{x^2} + {n^2}x + {{n\left( {{n^2} - 1} \right)} \over 3} = 10n$$

$$ \Rightarrow $$ $${x^2} + nx + {{\left( {{n^2} - 31} \right)} \over 3} = 0$$

Let $$\alpha $$ and $$\alpha $$ + 1 be its two solutions

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

$$ \Rightarrow $$ $$\alpha $$ = $${{ - n - 1} \over 2}$$ ....(1)

Also $$\alpha (\[\alpha $$ + 1) = $${{\left( {{n^2} - 31} \right)} \over 3}$$ ......(2)

Putting value of (1) in (2), we get

$$ - \left( {{{n + 1} \over 2}} \right)\left( {{{1 - n} \over 2}} \right) = {{\left( {{n^2} - 31} \right)} \over 3}$$

$$ \Rightarrow $$ n2 = 121

$$ \Rightarrow $$ n = 11

Solutions · 1