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Sum to 20 terms of the series (1.3)2 + (2.5)2 + (3.7)2 +… is:
- a)168090
- b)198090
- c)178090
- d)188090
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Sum to 20 terms of the series (1.3)2+ (2.5)2+ (3.7)2+… is:a)168...
Given that (1.3)2, (2.5)2 , (3.7)2 .....................20 terms
Here 1,2,3,.................are in A.P.
a = 1, d = 1
tn = a + (n-1) d = 1 + (n -1)1 = n
3,5,7 ............. are in A.P
a = 3, d = 2
tn = a + (n-1) d = 3 + (n -1)2 = 2n + 1
∴ nth term = n(2n + 1)2 = 4n3 + 4n2 + n
Sum of n terms Sn = ∑ ( 4n3 + 4n2 + n ) = 4 { n2(n+1)2}/4 + 4 {n (n+1)(2n+1)} / 6 + n(n+1)/2
= { n2(n+1)2} + 2 {n (n+1)(2n+1)} / 3 + n(n+1)/2
Sum of 20 terms S20 = 400 × 441 + 40 × 41 × 7 + 10 ×21 = 176400 + 11480 + 210
= 188090.
Here 1,2,3,.................are in A.P.
a = 1, d = 1
tn = a + (n-1) d = 1 + (n -1)1 = n
3,5,7 ............. are in A.P
a = 3, d = 2
tn = a + (n-1) d = 3 + (n -1)2 = 2n + 1
∴ nth term = n(2n + 1)2 = 4n3 + 4n2 + n
Sum of n terms Sn = ∑ ( 4n3 + 4n2 + n ) = 4 { n2(n+1)2}/4 + 4 {n (n+1)(2n+1)} / 6 + n(n+1)/2
= { n2(n+1)2} + 2 {n (n+1)(2n+1)} / 3 + n(n+1)/2
Sum of 20 terms S20 = 400 × 441 + 40 × 41 × 7 + 10 ×21 = 176400 + 11480 + 210
= 188090.
Most Upvoted Answer
Sum to 20 terms of the series (1.3)2+ (2.5)2+ (3.7)2+… is:a)168...
First try to make the nth term of the series nth term=4n^3+4n^2+n now find the sum of this nth term... we get sum of nth term= n^2(n+1)^2+4n(n+1)(2n+1)/6+n now put n=20 you will. get the ans 188090
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Community Answer
Sum to 20 terms of the series (1.3)2+ (2.5)2+ (3.7)2+… is:a)168...
To find the sum of the series, we can write out the first 20 terms and then add them up.
(1.3)^2 = 1.69
(2.5)^2 = 6.25
(3.7)^2 = 13.69
Continuing this pattern, the first 20 terms of the series are:
1.69, 6.25, 13.69, ...
Now, we can find the sum of these terms:
1.69 + 6.25 + 13.69 + ... + T20
To find T20, we need to find the 20th term of the series. The general formula for the nth term is:
Tn = (n + 0.3)^2
So, T20 = (20 + 0.3)^2 = 20.3^2 = 412.09
Now, we can find the sum of the series:
1.69 + 6.25 + 13.69 + ... + 412.09
Using the formula for the sum of an arithmetic series:
Sn = (n/2)(first term + last term)
Sn = (20/2)(1.69 + 412.09)
Sn = 10(1.69 + 412.09)
Sn = 4,139.8
Therefore, the sum of the first 20 terms of the series is 4,139.8.
(1.3)^2 = 1.69
(2.5)^2 = 6.25
(3.7)^2 = 13.69
Continuing this pattern, the first 20 terms of the series are:
1.69, 6.25, 13.69, ...
Now, we can find the sum of these terms:
1.69 + 6.25 + 13.69 + ... + T20
To find T20, we need to find the 20th term of the series. The general formula for the nth term is:
Tn = (n + 0.3)^2
So, T20 = (20 + 0.3)^2 = 20.3^2 = 412.09
Now, we can find the sum of the series:
1.69 + 6.25 + 13.69 + ... + 412.09
Using the formula for the sum of an arithmetic series:
Sn = (n/2)(first term + last term)
Sn = (20/2)(1.69 + 412.09)
Sn = 10(1.69 + 412.09)
Sn = 4,139.8
Therefore, the sum of the first 20 terms of the series is 4,139.8.
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Sum to 20 terms of the series (1.3)2+ (2.5)2+ (3.7)2+… is:a)168090b)198090c)178090d)188090Correct answer is option 'D'. Can you explain this answer?
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