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Algebra Level 3

Find the value of x.


The answer is 0.5.

Bogdan Rusu by Bogdan Rusu , 27, Romania

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1 solution

Harish Sasikumar
Nov 9, 2015

rewriting 1=2-1, 2=3-1, 3=4-1 and so on n=n+1-1

L H S = 2 x + ( 2 − 1 ) 2 + 2 x + ( 3 − 1 ) 3 + 2 x + ( 4 − 1 ) 4 + . . . . . . 2 x + ( n − 1 ) n + 2 x + ( n + 1 ) − 1 n + 1 LHS =\frac{2x+(2-1)}{2}+\frac{2x+(3-1)}{3}+\frac{2x+(4-1)}{4}+......\frac{2x+(n-1)}{n}+\frac{2x+(n+1)-1}{n+1}

= ( 2 x − 1 ) [ 1 2 + 1 3 + 1 4 + . . . . . . . . + 1 n + 1 n + 1 ] + [ 2 2 + 3 3 + 4 4 + . . . . . . + n n + n + 1 n + 1 ] =(2x-1)[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........+\frac{1}{n}+\frac{1}{n+1}]+[\frac{2}{2}+\frac{3}{3}+\frac{4}{4}+......+\frac{n}{n}+\frac{n+1}{n+1}]

= ( 2 x − 1 ) [ 1 2 + 1 3 + 1 4 + . . . . . . . . + 1 n + 1 n + 1 ] + n =(2x-1)[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........+\frac{1}{n}+\frac{1}{n+1}]+n

for LHS to be n, 2x-1=0 i.e x= 0.5 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Good recognition of the factoring that allows us to simplify the expression.

Alternatively, we can recognize that we have a linear equation, and so there is exactly one solution, which by observation is 0.5.

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