Just H and S

Calculus Level 4

If H = ∑ n = 1 ∞ n 3 ( n − 1 ) ! H = \displaystyle \sum_{n=1}^{\infty} \frac{n^{3}}{(n - 1)!} and S = ∑ n = 1 8 n 1 + n 2 + n 4 S = \displaystyle \sum_{n=1}^{8} \frac{n}{1+n^{2}+n^{4}}

The value of H = A ⋅ e H = A\cdot e and value of S = B C S = \frac{B}{C} .Then find A + B + C A+B+C .


The answer is 124.

Mohit Gupta by Mohit Gupta , 20, India

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2 solutions

Bharath Sriraam
Feb 25, 2016

We have 2 options for H

Option 1: We multiply n up and down so we get, H = ∑ n = 1 ∞ n 4 n ! H=\sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 4 } }{ n! } }

Now we know that this is 15e.

Option 2: We write n 3 ( n − 1 ) ! = n ( n 2 − 1 ) + n ( n − 1 ) ! \frac { { n }^{ 3 } }{ (n-1)! } =\frac { n({ n }^{ 2 }-1)+n }{ (n-1)! }

n ( n 2 − 1 ) + n ( n − 1 ) ! = n ( n − 1 ) ( n + 1 ) ( n − 1 ) ! + n − 1 + 1 ( n − 1 ) ! \frac { n({ n }^{ 2 }-1)+n }{ (n-1)! } =\frac { n(n-1)(n+1) }{ (n-1)! } +\frac { n-1+1 }{ (n-1)! }

= n ( n − 2 + 3 ) ( n − 2 ) ! + 1 ( n − 2 ) ! + 1 ( n − 1 ) ! = n ( n − 2 ) + 3 n ( n − 2 ) ! + 2 e ( A f t e r s u m m i n g t h e m u p ) =\frac { n(n-2+3) }{ (n-2)! } +\frac { 1 }{ (n-2)! } +\frac { 1 }{ (n-1)! } =\frac { n(n-2)+3n }{ (n-2)! } +2e(After\quad summing\quad them\quad up)

= n − 3 + 3 ( n − 3 ) ! + 3 [ n − 2 + 2 ( n − 2 ) ! ] =\frac { n-3+3 }{ (n-3)! } +3\left[ \frac { n-2+2 }{ (n-2)! } \right] + 2e

= 1 ( n − 4 ) ! + 3 ( n − 3 ) ! + 3 ( n − 3 ) ! + 6 ( n − 2 ) ! + 2 e =\frac { 1 }{ (n-4)! } +\frac { 3 }{ (n-3)! } +\frac { 3 }{ (n-3)! } +\frac { 6 }{ (n-2)! } +2e

After summing them up you get 15e.

Now moving onto S it can be written as

n ( n 2 + n + 1 ) ( n 2 − n + 1 ) = 1 2 { 1 n 2 − n + 1 − 1 n 2 + n + 1 } \frac { n }{ ({ n }^{ 2 }+n+1)({ n }^{ 2 }-n+1) } =\frac { 1 }{ 2 } \left\{ \frac { 1 }{ { n }^{ 2 }-n+1 } -\frac { 1 }{ { n }^{ 2 }+n+1 } \right\}

On expanding the summation you get 1 2 { 1 1 − 1 3 + 1 3 − 1 7 + 1 7 . . . − 1 73 } = 36 73 \frac { 1 }{ 2 } \left\{ \frac { 1 }{ 1 } -\frac { 1 }{ 3 } +\frac { 1 }{ 3 } -\frac { 1 }{ 7 } +\frac { 1 }{ 7 } ...-\frac { 1 }{ 73 } \right\} =\frac { 36 }{ 73 }

Now you compare the numbers with A,B and C and you get A+B+C=15+36+73=124. That's it!

Do pardon me for not using the summation symbols, if you can't understand a step please let me know.

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H could also be obtained by differentiating e^(x) and multiplying by x ,again differentiating e^(x) and multipling by x and repeating until we get the coefficient of x in form n^(4)/n!.

aryan goyat - 5 years, 3 months ago

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That's a very neat approach!

Bharath Sriraam - 5 years, 3 months ago

Hi Bharath.. I think there is a little mistake writing the solution on the first line..I think that it would be n^4/n!..Thank you so much for your solution..I enjoyed it a lot. I just made this comment to warn you about that.. Cheers

Jose Sacramento - 5 years, 3 months ago

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Thank you very much!

Bharath Sriraam - 5 years, 3 months ago
Archit Agrawal
Feb 25, 2016

A=15, B=36 and C=73.

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