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1 2 + 2 3 + 3 4 + .... n terms =

Question

$1 \times 2 + 2 \times 3 + 3 \times 4 + .... n t er m s =$

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

\frac{n ( n + 1 ) ( n - 2 )}{3}

\frac{n ( n + 1 ) ( n + 2 )}{6}

\frac{n ( n - 1 ) ( n - 2 )}{6}

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1 \times 2 + 2 \times 3 + 3 \times 4 + .... n t er m s =

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Text solutionVerified

n^{t h}

term of sequence

= n (n + 1)

Then sum of

n

terms

= r = 1 \sum n n (n + 1) = r = 1 \sum n (n^{2} + n)

= r = 1 \sum n n^{2} + r = 1 \sum n n = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} + \frac{n ( n + 1 )}{2}

= \frac{n ( n + 1 ) ( 2 n + 1 ) + 3 n ( n + 1 )}{6}

= \frac{n ( n + 1 ) [ 2 n + 1 + 3 ]}{6}

= \frac{n ( n + 1 ) ( 2 n + 4 )}{6}

= \frac{n ( n + 1 ) ( n + 2 )}{3}

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