Question -
Answer -
(i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10
Let us evaluate
We can write it as:
5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 = (1×2×3×4×5×6×7×8×9×10)/(1×2×3×4)
= 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) … (2n)] / (1) (2) (3).. (n)
= (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) … (2n)] / [(2) (4) (6) … (2n)]
= [(1) (2) (3) (4) …(2n-1) (2n)] / 2n [(1) (2) (3) … (n)]
= (2n)! / 2n n!