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the sum of first 40 terms 1,6,7,13,20.........???????
give slon
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- sum for this fibonacci series is S(n)= S(n+2)-S(2)
so for finding the sum upto 40th term we have to find 42nd term
f(42)=1095815001 (using C program :p)
so S(40)= S(42)-S(2)
S(40)=1095815001-6
so ans is S(40)= 1095814995 - 11 years agoHelpfull: Yes(5) No(7)
- sry solution is not completed
Sn=n(a1+an)/2
=40(236)/2
=40x118
=4720
- 11 years agoHelpfull: Yes(4) No(12)
- sum for this fibonacci series is S(n)= f(n+2)-f(2)(u can check by taking smaller value of n)
so for finding the sum upto 40th term we have to find 42nd term
f(42)=1095815001 (using C program :p)
so S(40)= f(42)-f(2)
S(40)=1095815001-6
so ans is S(40)= 1095814995 - 11 years agoHelpfull: Yes(3) No(2)
- as 1+6+7 is sum upto 3rd tem n it is 5th term -6, similarly
as 1+6+7+13+20+33+53 is sum upto 7th term and is equal to 9th term 139-6=133
hence sum upto 52nd term is 54th term-6
using golden ratio we can solve this, bt to be more accurate we calculate a few more terms. 1,6,7,13,20,33,53,86,139,225,364,589,953,1542....
thus 54th term= 1542*(1.6180339)^40
hence sum upto 52nd term is 352849113676-6=352849113670 - 11 years agoHelpfull: Yes(3) No(1)
- see,
1-------sum upto 1st term
7-------sum upto 2nd term
14-------sum upto 3rd term
27-------sum upto 4th term
47-------sum upto 5th term
80-------sum upto 6th term
133-------sum upto 7tht term
219-------sum upto 8th term
see Sum uptoNth term=(N+2)th term-6 (for this series.)
so sum upto 52th term=54th term-6
1+6+7+13+20+33+53=sum upto 7th term= 9th term-6=139-6
NOW USING "GOLDEN RATIO" WE CAN FIND THAT
1,6,7,13,...........................,139,225,364,589,953,1542(we have taken here 14TH TERM for accuracy)
so 54th=1542(14th term)*(14th term/13th term)^40
as we considered here 14th terms so the power is (54-14)=40
so we get,
1542*(1.6180339)^40=352849113676
so sum upto 52th=54th - 6
=352849113676 - 6
=352849113670 - 11 years agoHelpfull: Yes(1) No(0)
- think so the question is mistaken it must be like 1,7,13,20........
then
First find the 40th term:
a40 = a1+(n-1)d
given a1=1
d=a2-a1
=7-1
=6
=> a40= 1+39(6)
= 1+234
= 235
Then find the sum:
Sn=(a1+an)/2
=(1+235)/2
= 236/2
= 118
- 11 years agoHelpfull: Yes(0) No(28)
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