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 are constants, then the value of s is equal to

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Correct answer: B. - 1 2

aligned Let S & = _ k=1 ^n k(k+1)(k-1) \ & = _ k=1 ^n k^3-k aligned We know that _ r=1 ^P r= P(P+1) 2 and _ r=1 ^P r^3= [ P(P+1) 2 ]^2 aligned S & = [ n(n+1) 2 ]^2- [ n(n+1) 2 ] \ & = n(n+1) 2 [ n(n+1) 2 -1 ] \ & = ( n^2+n 2 ) [ n^2+n-2 2 ] \ & = n^4 4 + n^3 2 - n^2 4 - n 2 ...(i) aligned Now, by comparing p n^4+q n^3+t n^2+s n with Eq. (i), S=- 1 2

BITSAT Mathematics Question (2024) — Solution

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