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The sum upto infinity of the series 4/7-5/72+4/73-5/74+……..is
- a)23/48
- b)25/48
- c)1/2
- d)None
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The sum upto infinity of the series 4/7-5/72+4/73-5/74+…&hellip...
To find the sum of an infinite series, we need to find the limit of the partial sums as the number of terms approaches infinity. Let's first write out the partial sums:
S1 = 4/7 - 5/72
S2 = 4/7 - 5/72 + 4/73 - 5/74
S3 = 4/7 - 5/72 + 4/73 - 5/74 + 4/75 - 5/76
...
Notice that each even term in the partial sum cancels out with the previous odd term, so we can simplify the partial sums to:
S1 = 4/7 - 5/72
S2 = 4/73 - 5/74
S3 = 4/147 - 5/148
...
We can see that each term in the partial sum has the same form:
an = 4/(n*3-2) - 5/(n*3-1)
So, the nth partial sum is:
Sn = an + an-1 + an-2 + ... + a2 + a1
Using the formula for the sum of a geometric series, we can simplify this to:
Sn = (4/3) * [(1/(n*3-2)) - (1/7)]
- (5/3) * [(1/(n*3-1)) - (1/6)]
Now, taking the limit as n approaches infinity, we get:
lim Sn = (4/3) * [(1/3) - (1/7)]
- (5/3) * [(1/4) - (1/6)]
= 4/21 - 5/18
= -1/126
Therefore, the sum of the series is -1/126.
S1 = 4/7 - 5/72
S2 = 4/7 - 5/72 + 4/73 - 5/74
S3 = 4/7 - 5/72 + 4/73 - 5/74 + 4/75 - 5/76
...
Notice that each even term in the partial sum cancels out with the previous odd term, so we can simplify the partial sums to:
S1 = 4/7 - 5/72
S2 = 4/73 - 5/74
S3 = 4/147 - 5/148
...
We can see that each term in the partial sum has the same form:
an = 4/(n*3-2) - 5/(n*3-1)
So, the nth partial sum is:
Sn = an + an-1 + an-2 + ... + a2 + a1
Using the formula for the sum of a geometric series, we can simplify this to:
Sn = (4/3) * [(1/(n*3-2)) - (1/7)]
- (5/3) * [(1/(n*3-1)) - (1/6)]
Now, taking the limit as n approaches infinity, we get:
lim Sn = (4/3) * [(1/3) - (1/7)]
- (5/3) * [(1/4) - (1/6)]
= 4/21 - 5/18
= -1/126
Therefore, the sum of the series is -1/126.
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