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Q. The value of $\frac{1 \cdot 2 \cdot 3 \cdot 4+2 \cdot 3 \cdot 4 \cdot 5+3 \cdot 4 \cdot 5 \cdot 6+\ldots \ldots \text { upto } 16 \text { terms }}{1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\ldots \ldots \text { upto } 16 \text { terms }}$ is equal to

Sequences and Series

Solution:

$\Theta \frac{1 \cdot 2 \cdot 3 \cdot 4+2 \cdot 3 \cdot 4 \cdot 5+3 \cdot 4 \cdot 5 \cdot 6+\ldots \ldots . \text { upto } n \text { terms }}{1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\ldots . . . \text { upto } n \text { terms }}$
$=\frac{\frac{ n ( n +1)( n +2)( n +3)( n +4)}{5}}{\frac{ n ( n +1)( n +2)( n +3)}{4}}=\frac{4( n +4)}{5}$
For $n =16$, expression $=\frac{4 \times 20}{5}=16$