(i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10
Let us evaluate
We can write it as:
5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10
= \(\dfrac{ (1×2×3×4×5×6×7×8×9×10)}{(1×2×3×4)}\)
= \(\dfrac{ 10!}{4!}\)
(ii) 3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18
Let us evaluate
3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18 = (3×1) × (3×2) × (3×3) × (3×4) × (3×5) × (3×6)
= 36 (1×2×3×4×5×6)
= 36 (6!)
(iii) (n + 1) (n + 2) (n + 3) … (2n)
Let us evaluate
(n + 1) (n + 2) (n + 3) … (2n) = [(1) (2) (3) .. (n) … (n + 1) (n + 2) (n + 3) …
\(\dfrac{ (2n) }{ (1) (2) (3) .. (n)}\)
= \(\dfrac{ (2n)!}{n!}\)
(iv) 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1)
Let us evaluate
1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1) = [(1) (3) (5) … (2n-1)] [(2) (4) (6) …
\(\dfrac{ (2n) }{ (2) (4) (6) … (2n)}\)
= [(1) (2) (3) (4) … \(\dfrac{ (2n-1) (2n)] }{2n [(1) (2) (3) … (n)]}\)
=\(\dfrac{ (2n)! }{ 2^n n!}\)
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