find sum of n terms of the sequence 1.3.(2^2) + 2.4.(3^2)+3.5.(4^2)+.......

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The brute force way: $$\sum_{k=1}^nk(k+1)^2(k+2)=\sum_{k=1}^n(k^4+4k^3+5k^2+2k)=$$ $$=\tfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}+4\cdot\frac{n^2(n+1)^2}{4}+5\cdot\frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}{2}=$$ $$=\frac{n(n+1)(n+2)(n+3)(2n+3)}{10}.$$ Now, easy to make the telescopic sum. $$\sum_{k=1}^nk(k+1)^2(k+2)=$$ $$=\sum_{k=1}^n\left(\frac{k(k+1)(k+2)(k+3)(2k+3)}{10}-\frac{(k-1)k(k+1)(k+2)(2k+1)}{10}\right)=$$ $$=\frac{n(n+1)(n+2)(n+3)(2n+3)}{10}.$$

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Updated on November 04, 2022

Comments

  • Admin
    Admin about 1 year

    S=$1.3.2^2+2.4.3^2+3.5.4^2+........$ upto n terms

    Find S?

    This was asked under the heading using method of difference and ans given was (n)(n+1)(n+2)(n+3)(2n+3)/10

  • Hema
    Hema about 5 years
    I'm really sorry to trouble you after so long regarding this answer,I actually have a doubt, how exactly did you know which two terms to take for the telescopic sum? I have been struggling with such questions for quite a while. Please help me Sir.
  • Michael Rozenberg
    Michael Rozenberg about 5 years
    @Hema If we know the answer and what is it the telescoping summation, so just see my solution. $k(k+1)(k+2)$ it's the same factor and the rest must turn out.
  • Hema
    Hema about 5 years
    So it is by trial and error that we find the other factors? Is there any way to get a hint as to what they are, for instance that 2k+ something is factor and not 3k+something?
  • Michael Rozenberg
    Michael Rozenberg about 5 years
    I think this idea (the telescoping summation) does not always work. Here we are just lucky.