2014/095) Prove the sum of any 10 consecutive terms of the Fibonacci sequence is multiple of 11?

proof:

this can be proved as below
$F_3 = F_2 + F_1$
$F_4 = F_3 + F_2=2F_2+F_1$
$F_5 = F_4 + F_3=3F_2+2F_1$
$F_6 = F_5 + F_4=5F_2+3F_1$
$F_7 = F_6 + F_5=8F_2+5F_1$
$F_8 = F_7 + F_6=13F_2+8F_1$
$F_9 = F_8 + F_7=21F_2+13F_1$
$F_{10} = F_9 + F_8=34F_2+21F_1$

Adding we get $F_1+F_2+F_3+F_4+F_5+F_6+F_7+F_8+F_9+F_{10}$
=$88F_2+55F_1= 11(8F_2+5F_1) =11F_7$

This can be proved further below as(more fun)

$F_1+F_2+F_3+F_4+F_5+F_6+F_7+F_8+F_9+F_{10}$
=$(F_1+F_2)+(F_3+F_4)+(F_5+F_6)+F_7+F_8+F_9+F_{10}$
= $F_3+F_5+F_7+F_7+F_8+F_9+(F_8+F_9)$
= $F_3+F_5+2F_7+2F_8+2F_9$
= $F_3+F_5+2F_7+2F_8+2(F_7 + F_8)$
= $F_3+F_5+4F_7+4F_8$
=  $F_3+F_5+4F_7+4(F_6+F_7)$
=  $F_3+F_5+8F_7+4F_6$
=  $F_3+(F_5+F_6) +8F_7+3F_6$
= $F_3+9F_7+3F_6$
= $F_3+F_6+ 9F_7+2F_6$
=$F_3+(F_4+F_5)+ 9F_7+2F_6$
= $(F_3+F_4)+F_5+ 9F_7+2F_6$ 
= $F_5+F_5+ 9F_7+2F_6$
= $2 F_5+ 9F_7+2F_6$
= $2(F_5+F_6)+ 9F_7$
= $2F_7 + 9F_7$
= $11F_7$
Done