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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2024 · Mathematics · Sequences and Series

If ∑_ k=1 ^n k(k+1)(k-1)=p n^4+q n^3+t n^2+s n where p, q, t and s...

If ∑k=1nk(k+1)(k−1)=pn4+qn3+tn2+sn\sum_{k=1}^n k(k+1)(k-1)=p n^4+q n^3+t n^2+s n where p,q,tp, q, t and ss are constants, then the value of ss is equal to

Options

  1. A−14-\frac{1}{4}
  2. B−12-\frac{1}{2}
  3. C12\frac{1}{2}
  4. D14\frac{1}{4}

Correct answer

B. −12-\frac{1}{2}

Step-by-step solution

 Let S=∑k=1nk(k+1)(k−1)=∑k=1nk3−k\begin{aligned} \text{ Let } S & =\sum_{k=1}^n k(k+1)(k-1) \\ & =\sum_{k=1}^n k^3-k \end{aligned} We know that ∑r=1Pr=P(P+1)2\sum_{r=1}^P r=\frac{P(P+1)}{2} and ∑r=1Pr3=[P(P+1)2]2\sum_{r=1}^P r^3=\left[\frac{P(P+1)}{2}\right]^2S=[n(n+1)2]2−[n(n+1)2]=n(n+1)2[n(n+1)2−1]=(n2+n2)[n2+n−22]=n44+n32−n24−n2...(i)\begin{aligned} S & =\left[\frac{n(n+1)}{2}\right]^2-\left[\frac{n(n+1)}{2}\right] \\ & =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}-1\right] \\ & =\left(\frac{n^2+n}{2}\right)\left[\frac{n^2+n-2}{2}\right] \\ & =\frac{n^4}{4}+\frac{n^3}{2}-\frac{n^2}{4}-\frac{n}{2}...(i) \end{aligned}$Now,bycomparingNow, by comparingp n^4+q n^3+t n^2+s nwithEq.(i),with Eq. (i),S=-\frac{1}{2}$
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