Let a n = k = 1 ∑ n ( k + 2 ) j = 1 ∑ k i = 1 ∑ j i ! ( i 2 + i + 1 ) and b n = a n + 1 − a n . Compute b 4 .
Notation:
!
is the
factorial
notation. For example,
8
!
=
1
×
2
×
3
×
⋯
×
8
.
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a n + 1 = k = 1 ∑ n + 1 ( k + 2 ) j = 1 ∑ k i = 1 ∑ j i ! ( i 2 + i + 1 ) = [ { ( n + 1 ) + 2 } j = 1 ∑ n + 1 i = 1 ∑ j i ! ( i 2 + i + 1 ) ] + [ k = 1 ∑ n ( k + 2 ) j = 1 ∑ k i = 1 ∑ j i ! ( i 2 + i + 1 ) ] = ( n + 3 ) j = 1 ∑ n + 1 i = 1 ∑ j i ! ( i 2 + i + 1 ) + a n ∴ b n = a n + 1 − a n = ( n + 3 ) j = 1 ∑ n + 1 i = 1 ∑ j i ! ( i 2 + i + 1 )
Thus
b 4 = 7 j = 1 ∑ 5 i = 1 ∑ j i ! ( i 2 + i + 1 ) = 3 5 2 3 1
The calculations for the sum in b 4 :
j = 1 ∑ 5 i = 1 ∑ j i ! ( i 2 + i + 1 ) = j = 1 ∑ 5 [ i = 1 ∑ j i 2 ⋅ i ! + i = 1 ∑ j i ⋅ i ! + i = 1 ∑ j i ! ] = j = 1 ∑ 5 [ i = 1 ∑ j { ( i + 1 ) − 1 } 2 ⋅ i ! + i = 1 ∑ j { ( i + 1 ) ! − i ! } + i = 1 ∑ j i ! ] = j = 1 ∑ 5 [ i = 1 ∑ j ( i + 1 ) 2 ⋅ i ! − 2 i = 1 ∑ j ( i + 1 ) ⋅ i ! + i = 1 ∑ j i ! + i = 1 ∑ j ( i + 1 ) ! − i = 1 ∑ j i ! + i = 1 ∑ j i ! ] = j = 1 ∑ 5 [ i = 1 ∑ j ( i + 1 ) ⋅ ( i + 1 ) ! − 2 i = 1 ∑ j i ⋅ i ! − 2 i = 1 ∑ j i ! + i = 1 ∑ j i ! + i = 1 ∑ j ( i + 1 ) ! ] = j = 1 ∑ 5 [ i = 1 ∑ j { ( i + 2 ) ! − ( i + 1 ) ! } − 2 i = 1 ∑ j { ( i + 1 ) ! − i ! } − i = 1 ∑ j i ! + i = 1 ∑ j ( i + 1 ) ! ] = j = 1 ∑ 5 [ i = 1 ∑ j ( i + 2 ) ! − i = 1 ∑ j ( i + 1 ) ! − 2 i = 1 ∑ j ( i + 1 ) ! + 2 i = 1 ∑ j i ! − i = 1 ∑ j i ! + i = 1 ∑ j ( i + 1 ) ! ] = j = 1 ∑ 5 [ i = 1 ∑ j ( i + 2 ) ! − 2 i = 1 ∑ j ( i + 1 ) ! + i = 1 ∑ j i ! ] = j = 1 ∑ 5 [ i = 1 ∑ j { ( i + 2 ) ! − ( i + 1 ) ! } − i = 1 ∑ j { ( i + 1 ) ! − i ! } ] = j = 1 ∑ 5 [ { ( j + 2 ) ! − 2 ! } − { ( j + 1 ) ! − 1 ! } ] = j = 1 ∑ 5 { ( j + 2 ) ! − ( j + 1 ) ! } − j = 1 ∑ 5 { 2 ! − 1 ! } = ( 7 ! − 2 ! ) − 5 = 5 0 3 3
⟹ b 4 = 7 × 5 0 3 3 = 3 5 2 3 1
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Let π ( k ) π ( k ) a n b n b 4 = ( k + 2 ) j = 1 ∑ k i = 1 ∑ j i ! ( i 2 + i + 1 ) = ( k + 2 ) j = 1 ∑ k i = 1 ∑ j i ! ( ( i + 1 ) 2 − i ) = ( k + 2 ) j = 1 ∑ k i = 1 ∑ j ( i + 1 ) ( i + 1 ) ! − i ( i ! ) = ( k + 2 ) j = 1 ∑ k ( ( j + 1 ) ( j + 1 ) ! − 1 ) = ( k + 2 ) ( j = 1 ∑ k ( j + 1 ) ! ( j + 2 − 1 ) ) − k = ( k + 2 ) ( j = 1 ∑ k ( j + 2 ) ! − ( j + 1 ) ! ) − k = ( k + 2 ) ( ( ( k + 2 ) ! − 2 ) − k ) = ( k + 2 ) ( k + 2 ) ! − ( k + 2 ) 2 = k = 1 ∑ n π ( k ) = a n + 1 − a n = k = 1 ∑ n + 1 π ( k ) − k = 1 ∑ n π ( k ) = π ( n + 1 ) Thus we have, = π ( 5 ) = 7 ( 7 ! − 7 ) = 3 5 2 3 1 We can see that the sum is telescoping Once again, the sum is telescoping