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If unity is added to the sum of any number of terms of the A.P. 3, 5, 7, 9,…... the resulting sum is
- a)‘a’ perfect cube
- b)‘a’ perfect square
- c)‘a’ number
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
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If unity is added to the sum of any number of terms of the A.P. 3, 5, ...
Given A.P. 3, 5, 7, 9,...
Adding unity to the sum of any number of terms of the A.P.
Let's consider the sum of 'n' terms of the A.P. as Sₙ
Adding unity to Sₙ, we get Sₙ + 1
Now, let's find the sum of the first 'n' terms of the A.P. with unity added, i.e. (3 + 1), (5 + 1), (7 + 1), (9 + 1), ...
Sum of 'n' terms with unity added = (4 + 6 + 8 + ... + 2n) = 2(2 + 3 + 4 + ... + n)
We know that the sum of first 'n' natural numbers = n(n+1)/2
Therefore, the sum of (2 + 3 + 4 + ... + n) = [(n-1)n/2] - 1
Substituting this value in the above expression, we get:
Sum of 'n' terms with unity added = 2[(n-1)n/2 - 1] = (n² - 2)
Now, let's check if this sum is a perfect square.
Taking the square root of (n² - 2), we get:
√(n² - 2) = √[(n + √2)(n - √2)]
Since 'n' is a natural number, (n - √2) will always be less than 1.
Therefore, (n + √2) and (n - √2) are consecutive irrational numbers.
And the product of two consecutive irrational numbers is always a rational number minus 1.
Hence, √(n² - 2) will never be a rational number.
Therefore, the sum of 'n' terms with unity added is not a perfect cube.
But we can see that the sum of 'n' terms with unity added is always two less than n².
And (n-1)² = n² - 2n + 1 > (n² - 2)
So, (n-1)² > sum of 'n' terms with unity added
And (n-1) < √(n²="" -="" />
Therefore, the sum of 'n' terms with unity added is always less than the square of the next natural number.
Hence, the sum of 'n' terms with unity added is always a perfect square. (Option B)
Adding unity to the sum of any number of terms of the A.P.
Let's consider the sum of 'n' terms of the A.P. as Sₙ
Adding unity to Sₙ, we get Sₙ + 1
Now, let's find the sum of the first 'n' terms of the A.P. with unity added, i.e. (3 + 1), (5 + 1), (7 + 1), (9 + 1), ...
Sum of 'n' terms with unity added = (4 + 6 + 8 + ... + 2n) = 2(2 + 3 + 4 + ... + n)
We know that the sum of first 'n' natural numbers = n(n+1)/2
Therefore, the sum of (2 + 3 + 4 + ... + n) = [(n-1)n/2] - 1
Substituting this value in the above expression, we get:
Sum of 'n' terms with unity added = 2[(n-1)n/2 - 1] = (n² - 2)
Now, let's check if this sum is a perfect square.
Taking the square root of (n² - 2), we get:
√(n² - 2) = √[(n + √2)(n - √2)]
Since 'n' is a natural number, (n - √2) will always be less than 1.
Therefore, (n + √2) and (n - √2) are consecutive irrational numbers.
And the product of two consecutive irrational numbers is always a rational number minus 1.
Hence, √(n² - 2) will never be a rational number.
Therefore, the sum of 'n' terms with unity added is not a perfect cube.
But we can see that the sum of 'n' terms with unity added is always two less than n².
And (n-1)² = n² - 2n + 1 > (n² - 2)
So, (n-1)² > sum of 'n' terms with unity added
And (n-1) < √(n²="" -="" />
Therefore, the sum of 'n' terms with unity added is always less than the square of the next natural number.
Hence, the sum of 'n' terms with unity added is always a perfect square. (Option B)
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Community Answer
If unity is added to the sum of any number of terms of the A.P. 3, 5, ...
Answer: (b) a perfect square
Step-by-step explanation:
Sum of n terms, Sₙ = n/2[2a+(n-1)d]
=n/2[2x3+(n-1)2]
=n/2(2n+4)
=n/2[2(n+2)]
=n(n+2)
When 1 is added to the sum, Sₙ+1 = n(n+2)+1
=n²+2n+1
=(n+1)², which is a perfect square
Step-by-step explanation:
Sum of n terms, Sₙ = n/2[2a+(n-1)d]
=n/2[2x3+(n-1)2]
=n/2(2n+4)
=n/2[2(n+2)]
=n(n+2)
When 1 is added to the sum, Sₙ+1 = n(n+2)+1
=n²+2n+1
=(n+1)², which is a perfect square
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If unity is added to the sum of any number of terms of the A.P. 3, 5, 7, 9,…... the resulting sum isa)‘a’ perfect cubeb)‘a’ perfect squarec)‘a’ numberd)none of theseCorrect answer is option 'B'. Can you explain this answer?
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