# Let `u_n` =1.n+2.(n-1)+..........+(n-1).2+n.1 for any positive integer n. Prove, by the Principle of Mathematical Induction, that; `u_n` `=1/6 n (n+1)(n+2)` .``

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For mathematical induction we do three steps.

- Result true for n = 1
- For n = p where p>1 assume that result is true
- For n = p+1 show that the result is true.

`U_n = 1*n+2*(n-1)+..........+(n-1)*2+n*1---(1)`

`U_n = 1/6n(n+1)(n+2)`

When n=1

`U_1 = 1/6*1(1+1)(1+2) = 1`

From (1) it gives the first term as `U_1 = 1*1 = 1`

So n = 1 result is true.

Let us say for n = p where p>1 the result is true.

`U_p = 1/6p(p+1)(p+2)`

when n = p+1

We have to prove that `U_(p+1) = 1/6(p+1)(p+2)(p+3)`

From (1);

`U_n = 1*n+2*(n-1)+..........+(n-1)*2+n*1`

`U_(p+1) = 1*(p+1)+2(p)+3*(p-1)...........(p-1)*3+p*2+(p+1)*1`

`U_(p+1) = (1*p+1)+(2(p-1)+2)+(3(p-2)+3)....((p-1)*2+(p-1))+(p*1+p)+(p+1)`

`U_(p+1) = 1*p+2(p-1)+3(p-2)+......(p-1)*2+p*1+(1+2+3+...p+(p+1))`

`U_p+1 = U_p+(1+2+3+...p+(p+1))`

But we know that;

`(1+2+3+...p+(p+1)) = 1/2(p+1)(p+2)`

`U_p+1 = 1/6p(p+1)(p+2)+1/2(p+1)(p+2)`

`U_p+1 = 1/6(p+1)(p+2)(p+6/2)`

`U_p+1 = 1/6(p+1)(p+2)(p+3)`

So the result is true for `n = p+1`

** So from mathematical induction **`U_n = 1/6n(n+1)(n+2)`

**Sources:**