Convert the following products into factorials:
(i) 5 · 6 · 7 · 8 · 9 · 10
(ii) 3 · 6 · 9 · 12 · 15 · 18
(iii) (n + 1) (n + 2) (n + 3) ... (2n)
(iv) 1 · 3 · 5 · 7 · 9 ... (2n − 1)

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Solution

$(i) 5 \times 6 \times 7 \times 8 \times 9 \times 10 = \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10}{1 \times 2 \times 3 \times 4} = \frac{10!}{4!}$

$(ii) 3 \times 6 \times 9 \times 12 \times 15 \times 18 = (3 \times 1) \times (3 \times 2) \times (3 \times 3) \times (3 \times 4) \times (3 \times 5) \times (3 \times 6) = 3^{6} (1 \times 2 \times 3 \times 4 \times 5 \times 6) = 3^{6} (6!)$

$(iii) (n + 1) (n + 2) (n + 3) . . . (2 n) = \frac{(1) (2) (3) . . . (n) (n + 1) (n + 2) (n + 3) . . . (2 n)}{(1) (2) (3) . . . (n)} = \frac{(2 n)}{n!}$

$(iv) (1) (3) (5) . . . . . . . . . (2 n - 1) = \frac{[(1) (3) (5) . . . . . . . . . (2 n - 1)] [(2) (4) (6) . . . . . . . . . (2 n)]}{[(2) (4) (6) . . . . . . . . . (2 n)]} = \frac{[(1) (2) (3) (4) (5) . . . . . . . . . (2 n - 1) (2 n)]}{2^{n} [(1) (2) (3) . . . . . . . . . (n)]} = \frac{(2 n)!}{2^{n} n!}$

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Similar questions

Convert the following products into factorials:
(i) 5.6.7.8.9.10 (ii) 3.6.9.12.15.18
(iii) (n+1)(n+2)(n+3) ...(2n) (iv) 1.3.5.7.9 ...(2n-1)

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

$(1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) \dots (1 + \frac{(2 n + 1)}{n^{2}}) = (n + 1)^{2}$

Q. If

\frac{1}{1! 9!} + \frac{1}{3! 7!} + \frac{1}{5! 5!} + \frac{1}{7! 3!} + \frac{1}{9! 1!} = \frac{2^{n}}{10!}

Then n

=

Re-arrange suitably and find the sum in each of the following :

(i) $\frac{11}{12} + \frac{- 17}{3} + \frac{11}{2} + \frac{- 25}{2}$

(ii) $\frac{- 6}{7} + \frac{- 5}{6} + \frac{- 4}{9} + \frac{- 15}{7}$

(iii) $\frac{3}{5} + \frac{7}{3} + \frac{9}{5} + \frac{- 13}{15} + \frac{- 7}{3}$

(iv) $\frac{4}{13} + \frac{- 5}{8} + \frac{- 8}{13} + \frac{9}{13}$

(v) $\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}$

(vi) $\frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{- 5}{16}$

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Convert the following products into factorials: (i) 5 · 6 · 7 · 8 · 9 · 10 (ii) 3 · 6 · 9 · 12 · 15 · 18 (iii) (n + 1) (n + 2) (n + 3) ... (2n) (iv) 1 · 3 · 5 · 7 · 9 ... (2n − 1)

Convert the following products into factorials:
(i) 5 · 6 · 7 · 8 · 9 · 10
(ii) 3 · 6 · 9 · 12 · 15 · 18
(iii) (n + 1) (n + 2) (n + 3) ... (2n)
(iv) 1 · 3 · 5 · 7 · 9 ... (2n − 1)