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Algebra Level 5

Comprehension :- If r t h r^{th} term of a series can be written as a r = f ( r ) − f ( r − 1 ) a_r = f(r)-f(r-1) ,then S n = ∑ r = 0 n a r = f ( n ) − f ( 0 ) S_n= \sum_{r=0}^n a_r =f(n)-f(0) and S ∞ = lim ⁡ n → ∞ S n , S_\infty =\lim\limits_{n \to \infty} S_n, then answer the question

If 3 ( 3 ! ) + 4 ( 4 ! ) + 5 ( 5 ! ) + . . . 50 3(3!)+4(4!)+5(5!)+...50 t e r m s terms = a ! − b =a!-b then a − b a-b is equal to.. .

45 47 48 46

Atul Shivam by Atul Shivam , 23, India

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2 solutions

Deepak Kumar
Dec 13, 2015

Hint:express r(r)! as (r+1-1)(r)! And split it along '-' e.g (3) 3!=(4-1) 3!=4*3!-3!=4!-3!

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n ∗ n ! = ( n + 1 ) ! − ( n ) ! S 50 = ∑ n = 3 52 n ∗ ( n ! ) = ∑ n = 3 52 ( n + 1 ) ! − ∑ n = 3 52 ( n ! ) S 50 = ∑ n = 4 53 ( n ! ) − ∑ n = 3 52 ( n ! ) S 50 = 53 ! − 3 ! = a ! − b ! a − b = 47. n*n!=(n+1)! - (n)!\\ \displaystyle ~S_{50}=\sum_{n=3 }^{52 }n* (n !)=\sum_{n=3 }^{52 } (n+1)! - \sum_{n=3 }^{52 }(n!)\\ \displaystyle ~S_{50}=\sum_{n=4 }^{53} (n!) - \sum_{n=3 }^{52 }(n!)\\ S_{50}=53! - 3 !=a! - b!\\ a - b=\huge ~~~~~~~\color{#D61F06}{47}.

No other expressions are required!!!!!

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