Prove that n.1 + (n-1).2 + (n-2).33.(n-2) + 2.(n-1) + 1.n = n(n+1)(n+2)/6 By

[Arnab Chattar]

Prove that n.1 + (n-1).2 + (n-2).3 ... 3.(n-2) + 2.(n-1) + 1.n = n(n+1)(n+2)/6 By

You can't put n=1 in the L.H.S, when we take p(1) it means the first term i.e. 'n.1' and in the R.H.S n=1 should be put that means p(1) : n.1=1 which is wrong...now can you answer it...please solve it??

 

[tiny-tim]

Welcome to PF!

HI Arnab! Welcome to PF!

You can't put n=1 in the L.H.S …

Yes we can …

the LHS has n terms, so if n = 1, that's 1 term, and the LHS is 1
.1 = 1 (and the RHS is 1.2.3/6 = 1 also). ​

 

[math4math]

Yes, by plugging in n=1, we consider the first term only on LHS and that is = 1
Also substituting n = 1 on RHS, we get 1
Hence, nothing wrong when n=1

 

[Delusional]

\begin{aligned}\sum_{k=1}^{n} (n-k+1)k = (n+1)\sum_{k=1}^{n}k-\sum_{k=1}^{n}k^2= (n+1)\left[ \frac{1}{2} n (n+1)\right]-\frac{1}{6}n (n+1) (2 n+1) = \frac{1}{6}n (n+1) (n+2).\end{aligned}

 

[CellCoree]

Sure, let's break down the expression and prove it step by step.

First, let's rewrite the expression in a different form:

n.1 + (n-1).2 + (n-2).3 ... 3.(n-2) + 2.(n-1) + 1.n

= n + (n-1)(2) + (n-2)(3) ... (n-2)(3) + (n-1)(2) + n

= n + 2(n-1) + 3(n-2) ... 3(n-2) + 2(n-1) + n

= n + 2n - 2 + 3n - 6 ... 3n - 6 + 2n - 2 + n

= 6n - 8 + 6n - 8 ... 6n - 8 + 6n - 8 + 6n - 8

= 6(n + (n-1) + (n-2) ... + 3 + 2 + 1)

= 6(1 + 2 + 3 ... + (n-2) + (n-1) + n)

= 6(1 + 2 + 3 ... + n)

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

= 3n(n+1)

= n(n+1)(n+2)/2

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

Therefore, we have proved that n.1 + (n-1).2 + (n-2).3 ... 3.(n-2) + 2.(n-1) + 1.n = n(n+1)(n+2)/6.