JOMO 6, Short 3

Algebra Level 4

Given a sequence that starts with a 1 a_1 and each term is defined by:

a n = n 3 − 4 n − 1 , a_n = n^3 - 4n -1,

what are the last 3 digits of the sum of the first 66 terms of this sequence?


The answer is 611.

View wiki
Aditya Raut by Aditya Raut , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tasmeem Reza
Jul 20, 2014

A s t h e s e q u e n c e s t a r t s f r o m a 1 , S U M o f i t s f i r s t 66 t e r m s w i l l b e ∑ n = 1 66 a n As\: the\: sequence\: starts\: from\: a_{1},\: SUM\: of\: its\: first\: 66\: terms\: will\: be\: \sum_{n=1}^{66}a_{n} N o w , Now, ∑ n = 1 66 a n = ∑ n = 1 66 ( n 3 − 4 n − 1 ) \sum_{n=1}^{66}a_{n}=\sum_{n=1}^{66}(n^{3}-4n-1) = ∑ n = 1 66 n 3 − 4 ∑ n = 1 66 n − ∑ n = 1 66 1 =\sum_{n=1}^{66}n^{3}-4\sum_{n=1}^{66}n-\sum_{n=1}^{66}1 = ∑ n = 1 66 n 3 − 4 ∑ n = 1 66 n − 66 =\sum_{n=1}^{66}n^{3}-4\sum_{n=1}^{66}n-66 = ( 66 × 67 2 ) 2 − 4 ( 66 × 67 2 ) − 66 =\left ( \frac{66\times67}{2} \right )^{2}-4\left ( \frac{66\times67}{2} \right )-66 = 4879611 [ A f t e r s o m e t e d i o u s m u l t i p l i c a t i o n ] =4879611\: [After\: some\: tedious\: multiplication] w h o s e l a s t t h r e e d i g i t s a r e 611 whose\: last\: three\: digits\: are\: \boxed{611}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes, that's the expected way....

Aditya Raut - 6 years, 10 months ago

I'd do emphasis on "tedious multiplication" hahaha. Great :D

Carlos David Nexans - 6 years, 10 months ago

From what I understand the sum of n cubes is ( n ( n + 1 ) 2 ) 2 { \left( \frac { n(n+1) }{ 2 } \right) }^{ 2 } . Could someone prove this? I tried it with different values of n, and it always worked, but I am looking for a proof. Thanks.

Shashank Rammoorthy - 6 years, 5 months ago

Log in to reply

P r o o f b y i n d u c t i o n . A c c o r d i n g t o t h e P r i n c i p l e o f M a t h e m a t i c a l I n d u c t i o n , A s t a t e m e n t a b o u t i n t e g e r s i s t r u e f o r a l l i n t e g e r s g r e a t e r t h a n o r e q u a l t o 1 i f 1. ) i t i s t r u e f o r i n t e g e r 1 , a n d 2. ) w h e n e v e r i t i s t r u e f o r a l t h e i n t e g e r s 1 , 2 , . . . , k , t h e n i t i s t r u e f o r t h e i n t e g e r k + 1. 1 3 + 2 3 + 3 3 + . . . + n 3 = n 2 ( n + 1 ) 2 4 w h e n n = 1 , 1 ( 4 ) 4 = 1 3 1 = 1 T h e r e f o r e , i t i s t r u e f o r t h e i n t e g e r 1. I f i t i s t r u e f o r i n t e g e r s 1 , 2 , 3 , . . . , k t h e n i t m u s t b e t r u e f o r i n t e g e r k + 1. A s s u m i n g t h a t t h e f o r m u l a i s t r u e , 1 3 + 2 3 + 3 3 + . . . + ( k + 1 ) 3 = k 2 ( k + 1 ) 2 4 + ( k + 1 ) 3 = k 2 ( k + 1 ) 2 + 4 ( k + 1 ) 3 4 = ( k + 1 ) 2 ( k 2 + 4 k + 4 ) 4 = ( k + 1 ) 2 ( k + 2 ) 2 4 = ( k + 1 ) 2 [ ( k + 1 ) + 1 ] 2 4 ( k + 1 ) 2 [ ( k + 1 ) + 1 ] 2 4 i s i n t h e f o r m n 2 ( n + 1 ) 2 4 w h e r e n = k + 1 , s o i t i s t r u e f o r i n t e g e r k + 1. C o n d i t i o n s 1 a n d 2 a r e f u l f i l l e d . ∴ 1 3 + 2 3 + 3 3 + . . . + n 3 = n 2 ( n + 1 ) 2 4 f o r n ∈ Z , n ≥ 1. Proof\quad by\quad induction.\\ According\quad to\quad the\quad Principle\quad of\quad Mathematical\quad Induction,\\ A\quad statement\quad about\quad integers\quad is\quad true\quad for\quad all\quad integers\quad \\ greater\quad than\quad or\quad equal\quad to\quad 1\quad if\\ 1.)\quad it\quad is\quad true\quad for\quad integer\quad 1,\quad and\\ 2.)\quad whenever\quad it\quad is\quad true\quad for\quad al\quad the\quad integers\quad 1,2,...,k,\\ then\quad it\quad is\quad true\quad for\quad the\quad integer\quad k+1.\\ { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+...+{ n }^{ 3 }=\frac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 4 } \\ when\quad n=1,\\ \frac { 1(4) }{ 4 } ={ 1 }^{ 3 }\\ 1=1\\ Therefore,\quad it\quad is\quad true\quad for\quad the\quad integer\quad 1.\\ \\ If\quad it\quad is\quad true\quad for\quad integers\quad 1,\quad 2,\quad 3,...,\quad k\\ then\quad it\quad must\quad be\quad true\quad for\quad integer\quad k+1.\\ Assuming\quad that\quad the\quad formula\quad is\quad true,\\ { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+...+(k+1)^{ 3 }=\frac { k^{ 2 }{ (k+1) }^{ 2 } }{ 4 } +(k+1)^{ 3 }\quad \\ =\frac { k^{ 2 }{ (k+1) }^{ 2 }+4{ (k+1) }^{ 3 } }{ 4 } =\frac { { (k+1) }^{ 2 }({ k }^{ 2 }+4k+4) }{ 4 } \\ =\frac { { (k+1) }^{ 2 }{ (k+2) }^{ 2 } }{ 4 } =\frac { { (k+1) }^{ 2 }[(k+1)+1]^{ 2 } }{ 4 } \\ \frac { { (k+1) }^{ 2 }[(k+1)+1]^{ 2 } }{ 4 } \quad is\quad in\quad the\quad form\frac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 4 } \\ where\quad n=k+1,\quad so\quad it\quad is\quad true\quad for\quad integer\quad k+1.\\ Conditions\quad 1\quad and\quad 2\quad are\quad fulfilled.\\ \therefore \quad { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+...+{ n }^{ 3 }=\frac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 4 } \quad for\quad n∈Z,\quad n\ge 1.\\ \\ \\ \\

Justin Tuazon - 6 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...