# What is x in (x+1)C1+(2x+2)C2+(3x+3)C3=28?

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### 1 Answer

You should use factorial form of formula of combinations such that:

`C_n^k = (n!)/(k!*(n - k)!)`

Hence, you need to put in this form all the terms of the sum such that:

`C_(x+1)^1 = ((x+1)!)/(1!(x+1-1)!)`

`C_(x+1)^1 = ((x+1)!)/(x!) =gt C_(x+1)^1 = ((x)!(x+1))/(x!)`

Reducing factorial terms yields:

`C_(x+1)^1 = (x+1)`

You need to write the factorial form of the term `C_(2x+2)^` 2:

`C_(2x+2)^2 = ((2x+2)!)/(2!(2x+2-2)!)`

`C_(2x+2)^2 = ((2x+2)!)/(2!(2x)!)`

`C_(2x+2)^2 = ((2x)!(2x+1)(2x+2))/(2!(2x)!)`

Reducing factorial terms yields:

`C_(2x+2)^2 = ((2x+1)(2x+2))/(2!)`

You need to write the factorial form of the term `C_(3x+3)^3` :

`C_(3x+3)^3 = ((3x+3)!)/(3!(3x+3-3)!)`

`C_(3x+3)^3 = ((3x)!(3x+1)(3x+2)(3x+3))/(3!(3x)!)`

`C_(3x+3)^3 = ((3x+1)(3x+2)(3x+3))/(3!)`

You need to write the equation using the final forms found such that:

`(x+1) + ((2x+1)(2x+2))/2 + ((3x+1)(3x+2)(3x+3))/(1*2*3) = 28`

`6(x+1) + 3((2x+1)(2x+2)) + ((3x+1)(3x+2)(3x+3)) = 6*28`

You need to factor out 2 in `(2x+2)` and 3 in `(3x+3) ` such that:

`6(x+1) + 3*2(2x+1)(x+1) + 3(3x+1)(3x+2)(x+1) = 168`

You need to factor out 3(x+1) such that:

`3(x+1)*(2 + 2(2x+1) + (3x+1)(3x+2)) = 168`

You need to divide by 3 such that:

`(x+1)*(2 + 2(2x+1) + (3x+1)(3x+2)) = 56`

`(x+1)*(2 + 4x + 2 + 9x^2 + 9x + 2) = 56`

`(x+1)*(9x^2 + 13x + 6) = 56`

`9x^3 + 13x^2 + 6x + 9x^2 + 13x + 6 - 56 = 0`

`9x^3 + 22x^2 + 19x - 50 = 0`

You should check if `x = 1` is a zero for the equation `9x^3 + 22x^2 + 19x - 50 = 0` such that:

`9*(1)^3 + 22*(1)^2 + 19*1 - 50 = 9+22+19 - 50 = 50-50 = 0`

Hence, you may write the factored form of the polynomial `9x^3 + 22x^2 + 19x - 50 = 0` such that:

`9x^3 + 22x^2 + 19x - 50 = (x - 1)(ax^2 + bx + c)`

You need to open the brackets such that:

`9x^3 + 22x^2 + 19x - 50 = ax^3 + bx^2 + cx - ax^2 - bx - c`

Equating the coefficients of like terms such that:

`a = 9`

`b - a = 22 =gt b - 9 = 22 =gt b = 31`

`c - b = 19 =gt c - 31 = 19 =gt c = 50`

Hence, since `9x^3 + 22x^2 + 19x - 50 = 0 =gt (x - 1)(9x^2 + 31x + 50) = 0`

Notice that the quadratic expression is larger than 0 for all real x such that:

`9x^2 + 31x + 50 gt 0`

**Hence, the solution to the given equation is x = 1.**

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