_ n = 1 ^ 10 ( ( 6 _ i = 0 ^n i ) + 1 ( 6 _ j = 0 ^n (j - 1) ) + … - Vidyadip

MCQ

$\prod\limits_{n = 1}^{10} {\left( {\frac{{\left( {6\sum\limits_{i = 0}^n i } \right) + 1}}{{\left( {6\sum\limits_{j = 0}^n {(j - 1)} } \right) + 1}}} \right)} $ equals to

✓
$331$
B
$111$
C
$131$
D
$311$

✓

Answer

Correct option: A.

$331$

a
$\prod\limits_{n = 1}^{10} {\frac{{3{n^2} + 3n + 1}}{{3{n^2} - 3n + 1}} = } \frac{7}{1} \cdot \frac{{19}}{7} \cdot \frac{{37}}{{19}} \cdots \frac{{331}}{{271}} = 331$

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