Question

The sum $\displaystyle\sum^{10}_{r=1}(r^2 + 1) \times (r!)$ is equal to :

• A. $(11 !)$
• B. $10 \times (11!)$
• C. $101 \times (11!)$
• D. $11 \times (11!)$

Answer 1

The Correct Option is B

To solve the given problem, we need to evaluate the sum: $\displaystyle\sum^{10}_{r=1}(r^2 + 1) \times (r!)$.

1. First, expand the summation and calculate each term of the series: $(1^2 + 1) \times 1!$, $(2^2 + 1) \times 2!$, $(3^2 + 1) \times 3!$, ..., $(10^2 + 1) \times 10!$
2. Calculate each term individually:
   • $(1^2 + 1) \times 1! = 2 \times 1 = 2$
   • $(2^2 + 1) \times 2! = 5 \times 2 = 10$
   • $(3^2 + 1) \times 3! = 10 \times 6 = 60$
   • $(4^2 + 1) \times 4! = 17 \times 24 = 408$
   • $(5^2 + 1) \times 5! = 26 \times 120 = 3120$
   • $(6^2 + 1) \times 6! = 37 \times 720 = 26640$
   • $(7^2 + 1) \times 7! = 50 \times 5040 = 252000$
   • $(8^2 + 1) \times 8! = 65 \times 40320 = 2620800$
   • $(9^2 + 1) \times 9! = 82 \times 362880 = 29717760$
   • $(10^2 + 1) \times 10! = 101 \times 3628800 = 366508800$
3. Add all these terms to get the total sum:
   • $2 + 10 + 60 + 408 + 3120 + 26640 + 252000 + 2620800 + 29717760 + 366508800 \, = \, 403791360$
4. Now, factor $11!$ from the total sum:
   • $11! = 39916800$
5. Find the factor that will result in the original sum when multiplied by $11!$:
   • $403791360 \div 39916800 = 10$

Thus, the given sum $\displaystyle\sum^{10}_{r=1}(r^2 + 1) \times (r!)$ is equal to $10 \times (11!)$.

Therefore, the correct answer is $10 \times (11!)$.