Question

`\displaystyle\sum_( k = 1)^( n ) ( - 21 + 5k) = 996`Solve for "n" value.

Answer 1

Given an equation to solve for n:

`\sum ^n_(k\mathop=1)(-21+5k)=996`

The expansion of the given sum is as follows:

`\begin{gathered} -21+5(1)+(-21+5(2))+(-21+5(3))+\cdots+(-21+5n)=996 \\ -21n+(5+10+15+\cdots+5n)=996 \\ -21n+5(1+2+3+\cdots+n)=996 \\ -21n+5((n(n+1))/(2))=996 \\ -21n+(5)/(2)(n^2+n)=996 \\ -42n+5(n^2+n)=1992 \\ -42n+5n^2+5n=1992 \\ 5n^2-37n-1992=0 \end{gathered}`

Now, factorise the above quadratic equation:

`\begin{gathered} 5n^2+83n-120n-1992=0 \\ n(5n+83)-24(5n+83)=0 \\ (5n+83)(n-24)=0 \end{gathered}`

Use zero product rule in the equation to get:

5n + 83 = 0 or n - 24 = 0 which implies n = -83/5 and n = 24.

Neglect the negative solution of the equation to get n = 24.

Thus, the answer is 24.