# `1^4/(1cdot3)+2^4/(3cdot5)+...+n^4/((2n-1)(2n+1))=(n(n+1)*(n^2+n+1))/(6(2n+1))` prove using mathematical induction

First you need to check if the equality holds for `n=1.` ` `

`1^4/(1cdot3)=(1(1+1)(1^2+1+1))/(6(2cdot1+1))`

`1/3=1/3`

Now we assume that equality holds for all numbers less or equal to `n.`

And now we make inductive step in which we prove that equality holds for `n+1` as well.

`1^4/(1cdot3)+2^4/(3cdot5)+cdots+n^4/((2n-1)(2n+1))+((n+1)^4)/((2n+1)(2n+3))=`

By our assumption this equals to

`(n(n+1)(n^2+n+1))/(6(2n+1))+((n+1)^4)/((2n+1)(2n+3))=`

`((1 + n) (6 + 9 n + 5 n^2 + n^3))/(6 (3 + 2 n))=`

`((n+1)(n+2)(n^2+3n+3))/(6(2n+3))=`

`((n+1)((n+1)+1)((n+1)^2(n+1)+1))/(6(2(n+1)+1))`

This means that our equality holds for `n+1` which completes inductive step meaning we have proven the equality by mathematical induction.

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