the sum of 40 terms of the series 1+2+3+4+5+8+7+16+9+...is
Answers
The sum of 40 terms of the series 1 + 2 + 3 + 4 + 5 + 8 + 7 + 16 +9+..= + 398
Step-by-step explanation:
The given series has a combination of two series:
1 + 3 + 5 + 7 + . . . . . . (20 terms) are in AP and
2 + 4 + 8 + 16 + . . . . . . . . (20 terms) are in GP.
To find, the sum of 40 terms of the series 1 + 2 + 3 + 4 + 5 + 8 + 7 + 16 +9+.. = ?
1 + 3 + 5 + 7 + . . . . . . (20 terms) are in AP
Here, first term(a) = 1, common difference(d) = 3 - 1 = 2 and number of terms(n) = 20
= 400
2 + 4 + 8 + 16 + . . . . . . . . (20 terms) are in GP.
Here, first term(a) = 2, common ratio(r) = 2 and number of terms(n) = 20
The sum of 40 terms of the series 1 + 2 + 3 + 4 + 5 + 8 + 7 + 16 +9+..
= 400 + - 2
= + 398
Step-by-step explanation:
The sum of 40 terms of the series 1 + 2 + 3 + 4 + 5 + 8 + 7 + 16 +9+..= 2^{21}2
21
+ 398
Step-by-step explanation:
The given series has a combination of two series:
1 + 3 + 5 + 7 + . . . . . . (20 terms) are in AP and
2 + 4 + 8 + 16 + . . . . . . . . (20 terms) are in GP.
To find, the sum of 40 terms of the series 1 + 2 + 3 + 4 + 5 + 8 + 7 + 16 +9+.. = ?
1 + 3 + 5 + 7 + . . . . . . (20 terms) are in AP
Here, first term(a) = 1, common difference(d) = 3 - 1 = 2 and number of terms(n) = 20
S_{n}=\dfrac{n}{2}(2a+(n-1)d)S
n
=
2
n
(2a+(n−1)d)
S_{20}=\dfrac{20}{2}[2(1)+(20-1)2]S
20
=
2
20
[2(1)+(20−1)2]
=10(2+38)=10(2+38)
= 400
2 + 4 + 8 + 16 + . . . . . . . . (20 terms) are in GP.
Here, first term(a) = 2, common ratio(r) = 2 and number of terms(n) = 20
S_{n} =\dfrac{a(r^n-1)}{r-1}S
n
=
r−1
a(r
n
−1)
S_{20} =\dfrac{2(2^{20}-1)}{2-1}S
20
=
2−1
2(2
20
−1)
=2^{20+1}-2=2
20+1
−2
=2^{21}-2=2
21
−2
The sum of 40 terms of the series 1 + 2 + 3 + 4 + 5 + 8 + 7 + 16 +9+..
= 400 + 2^{21}2
21
- 2
= 2^{21}2
21
+ 398