The sum of the series 1 1 2 3 4 + 1 2 3 4 5 + n-terms is,

Mathematics · JEE Main · NTA Exams — Progression and Series

The sum of the series \(\frac{1}{1 \cdot 2 \cdot 3 \cdot 4}+\frac{1}{2 \cdot 3 \cdot 4 \cdot 5}+\ldots\) n-terms is,

\(\frac{1}{3}\left[\frac{1}{2 \cdot 1}-\frac{1}{n(n+2)(n+3)}\right]\)
\(\frac{1}{2}\left[\frac{1}{2 \cdot 1 \cdot 3}-\frac{1}{n(n+1)(n+2)}\right]\)
\(\frac{1}{3}\left[\frac{1}{2 \cdot 1 \cdot 3}-\frac{1}{n(n+1)(n+2)}\right]\)
\(\frac{1}{3}\left[\frac{1}{2 \cdot 1 \cdot 3}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)

Answer

(D) 1 3 [ 1 2 1 3 - 1 (n+1)(n+2)(n+3) ]

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