# Can anyone help solve equation 3!/1!+5!/3!+.......+(2n+1)!/(2n-1)!=406?

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You should write the general term of the sum as `a_k = ((2k+1)!)/((2k-1)!).`

You need to remember that you may write the factorial form such that:

`(2k+1)! = 1*2*3*...*(2k-1)*2k*(2k+1)`

Notice that you may substitute `(2k-1)!` for `1*2*3*...*(2k-1) ` such that:

(2k+1)! = (2k-1)!*2k*(2k+1)

Hence, `a_k = ((2k-1)!*2k*(2k+1))/((2k-1)!).`

`a_k = 2k*(2k+1) => a_k = 4k^2 + 2k`

You need to use the summations of polynomial expressions such that:

`sum_(k=1)^n a_k = sum_(k=1)^n (4k^2 + 2k)`

`sum_(k=1)^n a_k = sum_(k=1)^n 4k^2+ sum_(k=1)^n 2k`

`sum_(k=1)^n a_k = 4(n(n+1)(2n+1))/6 + 2n(n+1)/2`

`sum_(k=1)^n a_k = 2(n(n+1)(2n+1))/3 + n(n+1)`

Factoring out `n(n+1)` yields:

`sum_(k=1)^n a_k = n(n+1)(4n/3 + 2/3 + 1)`

You need to solve for n the equation `sum_(k=1)^n a_k = 406` such that:

`n(n+1)(4n/3 + 2/3 + 1) = 406`

`n(n+1)(4n + 5) = 1218`

`4n^3 + 5n^2 + 4n^2 + 5n - 1218 = 0`

`4n^3 + 9n^2 + 5n - 1218 = 0`

Notice that for `n = 6` , the equation holds such that:

`864 + 324 + 30 - 1218 = 0 => 1218 - 1218 = 0 => 0=0`

Notice that `n=6` is the only real root, the next two roots are complex conjugate.

**Hence, evaluating the natural root to this equation yields `n = 6.` **

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