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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

  1. Jul 17, 2011 #1
    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 cant 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...plz solve it??
     
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  3. Jul 17, 2011 #2

    tiny-tim

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    Welcome to PF!

    HI Arnab! Welcome to PF! :wink:
    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). :smile:
     
  4. Jul 17, 2011 #3
    Re: Prove that n.1 + (n-1).2 + (n-2).3 ....... 3.(n-2) + 2.(n-1) + 1.n = n(n+1)(n+2)/

    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
     
  5. Jul 19, 2011 #4
    Re: Prove that n.1 + (n-1).2 + (n-2).3 ....... 3.(n-2) + 2.(n-1) + 1.n = n(n+1)(n+2)/

    \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}
     
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