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Maths-

General

Easy

Question

The sum of the series $1 times 2 times 3 plus 2 times 3 times 4 plus 3 times 4 times 5 plus...$ to $n$ terms is

$n (n + 1) (n + 2)$
$(n + 1) (n + 2) (n + 3)$
$\frac{1}{4} n (n + 1) (n + 2) (n + 3)$
$\frac{1}{4} (n + 1) (n + 2) (n + 3)$

The correct answer is: $fraction numerator 1 over denominator 4 end fraction n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses left parenthesis n plus 3 right parenthesis$

Here, $T subscript n end subscript equals n open parentheses n plus 1 close parentheses left parenthesis n plus 2 right parenthesis$
$equals n to the power of 3 end exponent plus 3 n to the power of 2 end exponent plus 2 n$
$therefore blank S subscript n end subscript equals not stretchy sum T subscript n end subscript$
$equals open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets to the power of 3 end exponent plus fraction numerator 3 n open parentheses n plus 1 close parentheses left parenthesis 2 n plus 1 right parenthesis over denominator 6 end fraction plus fraction numerator 2 n left parenthesis n plus 1 right parenthesis over denominator 2 end fraction$
$equals fraction numerator 1 over denominator 4 end fraction n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses open parentheses n plus 3 close parentheses$

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