The value of the sum 1*2*3 + 2*3*4 + 3*4*5 + ... + n(n+1)(n+2) is:

A. n(n+1)(n+2)(n+3)4 — Using the identity k(k+1)(k+2) = [k(k+1)(k+2)(k+3) - (k-1)k(k+1)(k+2)]/4, the sum telescopes to the final term divided by 4.

JEE Advanced · MathematicsMedium

$The value of the sum 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots + n (n + 1) (n + 2) is:$

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Correct answer: A — $\frac{n ( n + 1 ) ( n + 2 ) ( n + 3 )}{4}$

Using the identity k (k + 1) (k + 2) = \frac{k ( k + 1 ) ( k + 2 ) ( k + 3 ) - ( k - 1 ) k ( k + 1 ) ( k + 2 )}{4}, the sum telescopes to \frac{n ( n + 1 ) ( n + 2 ) ( n + 3 )}{4} .