Find the floor

Algebra Level 4

If the general term of a series is T n = ( n 4 + 3 n 3 + 6 n 2 + 3 n + 1 ) n ! T_n=(n^4+3n^3+6n^2+3n+1)n! and if ∑ n = 1 30 T n = S \sum_{n=1}^{30} T_n=S then find ⌊ S 31 ∗ 1 0 36 ⌋ \large \lfloor \frac{S}{31*10^{36}} \rfloor


The answer is 7.

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Sanjeet Raria by Sanjeet Raria , 27, India

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1 solution

Ayush Verma
Nov 22, 2014

T n n ! = ( n 4 + 3 n 3 + 6 n 2 + 3 n + 1 ) = ( n + 4 ) ( n + 3 ) ( n + 2 ) ( n + 1 ) − 7 ( n + 3 ) ( n + 2 ) ( n + 1 ) + 13 ( n + 2 ) ( n + 1 ) − 9 ( n + 1 ) + 2 ⇒ T n = ( n + 4 ) ! − 7 ( n + 3 ) ! + 13 ( n + 2 ) ! − 9 ( n + 1 ) ! + 2 n ! = { ( n + 4 ) ! − ( n + 3 ) ! } − 6 { ( n + 3 ) ! − ( n + 2 ) ! } + 7 { ( n + 2 ) ! − ( n + 1 ) ! } − 2 { ( n + 1 ) ! − n ! } \cfrac { { T }_{ n } }{ n! } =\left( { n }^{ 4 }+3{ n }^{ 3 }+6{ n }^{ 2 }+3n+1 \right) \\ \\ \quad \quad \quad =\left( n+4 \right) \left( n+3 \right) \left( n+2 \right) \left( n+1 \right) -7\left( n+3 \right) \left( n+2 \right) \left( n+1 \right) +13\left( n+2 \right) \left( n+1 \right) -9\left( n+1 \right) +2\\ \\ \Rightarrow { T }_{ n }=\left( n+4 \right) !-7\left( n+3 \right) !+13\left( n+2 \right) !-9\left( n+1 \right) !+2n!\\ \\ \quad \quad =\left\{ \left( n+4 \right) !-\left( n+3 \right) ! \right\} -6\left\{ \left( n+3 \right) !-\left( n+2 \right) ! \right\} +7\left\{ \left( n+2 \right) !-\left( n+1 \right) ! \right\} -2\left\{ \left( n+1 \right) !-n! \right\}

this is telescoping,

∴ S = ( 34 ! − 4 ! ) − 6 ( 33 ! − 3 ! ) + 7 ( 32 ! − 2 ! ) − 2 ( 31 ! − 1 ! ) \therefore S=\left( 34!-4! \right) -6\left( 33!-3! \right) +7\left( 32!-2! \right) -2\left( 31!-1! \right)

Now use calculator,

S = 2.449583635 × 10 38 A n s = ⌊ 7.90188 ⌋ = 7 S=2.449583635\times { 10 }^{ 38 }\\ \\ Ans=\left\lfloor 7.90188 \right\rfloor =7

37 Helpful 0 Interesting 1 Brilliant 1 Confused

Another way of approaching this question could be to know that T n = [ ( ( n + 1 ) 3 − 1 ) ( n + 1 ) ! − ( n 3 − 1 ) n ! ] T_n=[((n+1)^3-1)(n+1)!-(n^3-1)n!] Now the series can easily be telescoped to get the result.

@Pranjal Jain @Ayush Verma @sandeep Rathod

Sanjeet Raria - 6 years, 6 months ago

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I like Ayush's method, as it easily generalizes and shows us how to deal with ∑ f ( n ) n ! \sum f(n) n! for some polynomial f ( n ) f(n) . It doesn't involve trying to spot a "magical" equation, but shows us how we can derive it.

Calvin Lin Staff - 6 years, 6 months ago

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Ayush's method is brilliant no doubt in that but the equation I've written down is not magical since everyone can reach it in a more generalized and simple way. I am going to explain this in my upcoming note.

Sanjeet Raria - 6 years, 6 months ago

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@Sanjeet Raria – I look forward to learning about that! It will help such telescoping sums to be able to quickly figure out how to get f ( n ) = g ( n + 1 ) − g ( n ) f(n) = g (n+1) - g(n) .

Calvin Lin Staff - 6 years, 6 months ago

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@Calvin Lin – Exactly sir.

Sanjeet Raria - 6 years, 6 months ago

( n 4 + 3 n 3 + 6 n 2 + 3 n + 1 ) n ! (n^4+3n^3+6n^2+3n+1)n!

= ( n 4 + 4 n 3 + 6 n 2 + 4 n + 1 − n 3 − n ) n ! =(n^4+4n^3+6n^2+4n+1-n^3-n)n!

= ( n + 1 ) 4 n ! − ( n 3 ) n ! − ( n ) n ! =(n+1)^4n!-(n^3)n!-(n)n!

= ( n + 1 ) 3 ( n + 1 ) n ! − ( n 3 ) n ! − ( n + 1 − 1 ) n ! =(n+1)^3(n+1)n!-(n^3)n!-(n+1-1)n!

= ( n + 1 ) 3 ( n + 1 ) ! − ( n 3 ) n ! − ( n + 1 ) n ! + n ! =(n+1)^3(n+1)!-(n^3)n!-(n+1)n!+n!

= ( ( n + 1 ) 3 ( n + 1 ) ! − ( n + 1 ) ! ) − ( ( n 3 ) n ! − n ! ) =((n+1)^3(n+1)!-(n+1)!)-((n^3)n!-n!)

= ( ( n + 1 ) 3 − 1 ) ( n + 1 ) ! ) − ( ( n 3 − 1 ) n ! ) =((n+1)^3-1)(n+1)!)-((n^3-1)n!)

g ( n ) = ( n 3 − 1 ) n ! g(n)=(n^3-1)n!

John Frank - 4 years, 9 months ago

How can someone think of telescoping here?? Awesome!!

Pranjal Jain - 6 years, 6 months ago

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That is quite natural in the context of this question. You may be familiar with the case of ∑ i × i ! = ( n + 1 ) ! − 1 \sum i \times i ! = (n+1) ! -1 . This is often done with induction, but the better approach is to use telescoping series and recognize it as ∑ i × i ! = ∑ [ ( i + 1 ) ! − i ! ] = ( n + 1 ) ! − 1 ! \sum i \times i! = \sum [ (i+1)! - i!] = (n+1) ! - 1! .

Hence, we want to do something similar. The easiest way to generalize it to to look at ∑ ( ( i + 2 ) ! − ( i + 1 ) ! = ∑ ( i + 1 ) 2 × i ! \sum ( (i+2)! - (i+1)! = \sum ( i+1) ^2 \times i! . This gives a way to motivate the rest of the proof.

Calvin Lin Staff - 6 years, 6 months ago

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Nicely explained! Thanks! You too @Ayush Verma

Pranjal Jain - 6 years, 6 months ago

Hats off . Upvoted

This image is copied from John Muradeli comment

U Z - 6 years, 6 months ago

I appreciate your talent. Outstanding.

Sanjeet Raria - 6 years, 6 months ago

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THANKS,SIR

Ayush Verma - 6 years, 6 months ago

Sir how did you factorized it? @Ayush Verma

sandeep Rathod - 6 years, 6 months ago

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Method 1

( n 4 + 3 n 3 + 6 n 2 + 3 n + 1 ) n ! = { n 3 ( n + 1 ) + 2 n 2 ( n + 1 ) + 4 n ( n + 1 ) − ( n + 1 ) + 2 } n ! = { n 2 ( n + 2 ) + 4 ( n + 2 ) − 9 } ( n + 1 ) ! + 2 n ! = { n ( n + 3 ) − 3 ( n + 3 ) + 13 } ( n + 2 ) ! − 9 ( n + 1 ) ! + 2 n ! = { ( n + 4 ) − 7 } ( n + 3 ) ! + 13 ( n + 2 ) ! − 9 ( n + 1 ) ! + 2 n ! = ( n + 4 ) ! − 7 ( n + 3 ) ! + 13 ( n + 2 ) ! − 9 ( n + 1 ) ! + 2 n ! \left( { n }^{ 4 }+{ 3n }^{ 3 }+6{ n }^{ 2 }+3n+1 \right) n!\\ \\ =\left\{ { n }^{ 3 }\left( n+1 \right) +2{ n }^{ 2 }\left( n+1 \right) +4n\left( n+1 \right) -\left( n+1 \right) +2 \right\} n!\\ \\ =\left\{ { n }^{ 2 }\left( n+2 \right) +4\left( n+2 \right) -9 \right\} \left( n+1 \right) !+2n!\\ \\ =\left\{ n\left( n+3 \right) -3\left( n+3 \right) +13 \right\} \left( n+2 \right) !-9\left( n+1 \right) !+2n!\\ \\ =\left\{ \left( n+4 \right) -7 \right\} \left( n+3 \right) !+13\left( n+2 \right) !-9\left( n+1 \right) !+2n!\\ \\ =\left( n+4 \right) !-7\left( n+3 \right) !+13\left( n+2 \right) !-9\left( n+1 \right) !+2n!

Method 2

( n 4 + 3 n 3 + 6 n 2 + 3 n + 1 ) = A ( n + 4 ) ( n + 3 ) ( n + 2 ) ( n + 1 ) + B ( n + 3 ) ( n + 2 ) ( n + 1 ) + C ( n + 2 ) ( n + 1 ) + D ( n + 1 ) + E C o m p a r e t h e c o e f f i c i e n t o f n 4 , n 3 , n 2 , n , 1 & s o l v e f o r A , B , C , D & E \left( { n }^{ 4 }+{ 3n }^{ 3 }+6{ n }^{ 2 }+3n+1 \right) \\ \\ =A\left( n+4 \right) \left( n+3 \right) \left( n+2 \right) \left( n+1 \right) \\ \\ +B\left( n+3 \right) \left( n+2 \right) \left( n+1 \right) \\ \\ +C\left( n+2 \right) \left( n+1 \right) \\ \\ +D\left( n+1 \right) +E\\ \\ Compare\quad the\quad coefficient\quad of\quad { n }^{ 4 },{ n }^{ 3 },{ n }^{ 2 },n,1\quad \\ \\ \& \quad solve\quad for\quad A,B,C,D\& E

Ayush Verma - 6 years, 6 months ago

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Really nice greatly done

sandeep Rathod - 6 years, 6 months ago

Actually It is 7.9018827

Shashank Rustagi - 5 years, 12 months ago

Killed it!!!!!! One of the best solutions I have seen over here...

Surya Sharma - 5 years, 3 months ago

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Thanks,Surya.

Ayush Verma - 5 years, 3 months ago

Floor{(1/(31×10^36))×{sum (n! (n^4+3n^3+6n^2+3n+1)), n=1 to 30}}

=7 WolframAlpha

Harout G. Vartanian - 4 years, 4 months ago

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