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Suppose you are asked to find the sum of the first ten natural numbers (\(1+2+3...+10\)). You could add them manually. But what if you are asked to sum the first 100, or the first 1,000 numbers? [span_2](start_span)[span_2](end_span)
By writing the sequence forwards and backwards, we can derive a brilliant shortcut. The sum of the first \(n\) natural numbers is exactly \(S_n = \frac{n(n+1)}{2}\). This result was documented centuries ago in ΔryabhaαΉa's ΔryabhaαΉΔ«ya! [span_3](start_span)[span_3](end_span) Let's explore how it works.
By writing the sequence forwards and backwards, we can derive a brilliant shortcut. The sum of the first \(n\) natural numbers is exactly \(S_n = \frac{n(n+1)}{2}\). This result was documented centuries ago in ΔryabhaαΉa's ΔryabhaαΉΔ«ya! [span_3](start_span)[span_3](end_span) Let's explore how it works.
1 The Reversing Sum Method
2 Using the Sum Formula
3 Summing Middle Chunks
Select the correct option:
What is the algebraic formula for finding the sum of the first \(n\) natural numbers (\(S_n\))? [span_18](start_span)[span_18](end_span)
Concept:
1
Because we added the forward sequence to the backward sequence, we generated \(n\) pairs of \((n+1)\). [span_19](start_span)[span_19](end_span)
2
Because we essentially calculated the sequence twice, we must divide by 2. [span_20](start_span)[span_20](end_span)
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Hint:
1
Look for the formula that multiplies \(n\) by the next consecutive number, then halves it.
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Mistakes to Avoid:
!
Selecting the full AP sum: While Option A works if you set \(a=1\) and \(d=1\), Option B is the specific, beautifully simplified formula for natural numbers.
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Solution:
As derived in the textbook, \(2S = n(n+1)\), which concludes with \(S_n = \frac{n(n+1)}{2}\). [span_21](start_span)[span_21](end_span)
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Select the correct option:
What is the sum of the first 10 natural numbers? [span_22](start_span)[span_22](end_span)
Concept:
1
Use \(S_n = \frac{n(n+1)}{2}\) and substitute \(n=10\). [span_23](start_span)[span_23](end_span)[span_24](start_span)[span_24](end_span)
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Hint:
1
Calculate \(10 \times 11\).
2
Divide the result by 2. [span_25](start_span)[span_25](end_span)
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Mistakes to Avoid:
!
Forgetting the denominator: If you stop at \(10 \times 11 = 110\), you are calculating \(2S\), not \(S\).
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Solution:
Using the formula for \(n=10\):
\(S_{10} = \frac{10(11)}{2}\)
\(S_{10} = 5 \times 11 = 55\) [span_26](start_span)[span_26](end_span)
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Select the correct option:
Which ancient mathematical text is historically credited with providing an early written mention of this summation formula? [span_27](start_span)[span_27](end_span)
Concept:
1
The formula \(S_n = \frac{n(n+1)}{2}\) is not a modern invention; its logic was elegantly summarized in ancient Indian texts. [span_28](start_span)[span_28](end_span)
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Hint:
1
Look for the text authored by ΔryabhaαΉa.
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Mistakes to Avoid:
!
Confusing mathematicians: The textbook credits VirahΔnka for the Fibonacci sequence (VrttajΔtisamuchaya) and ΔryabhaαΉa for the sum of natural numbers. [span_29](start_span)[span_29](end_span)[span_30](start_span)[span_30](end_span)
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Solution:
The textbook explicitly states: "The first known written mention of this result can be found in ΔryabhaαΉa's ΔryabhaαΉΔ«ya, Chapter 2, Verse 19." [span_31](start_span)[span_31](end_span)
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Select the correct option:
In the textbook's pictorial representation of the sum \(1+2+3+4+5+6\), what are the dimensions of the rectangular array of dots formed before halving it? [span_32](start_span)[span_32](end_span)
Concept:
1
A pictorial sum builds a triangle of dots. If you copy that triangle and flip it upside down next to the original, it forms a rectangle. [span_33](start_span)[span_33](end_span)
2
The rectangle has a height of \(n\) and a width of \((n+1)\). [span_34](start_span)[span_34](end_span)
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Hint:
1
Substitute \(n=6\) into the dimensions \(n\) and \((n+1)\).
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Mistakes to Avoid:
!
Assuming it makes a square: Only the sum of odd numbers makes a perfect square. The sum of natural numbers makes a slightly off-center rectangle.
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Solution:
As shown in Figure 8.5, combining two triangular sets of dots for \(n=6\) creates a rectangle.
The dimensions are \((n+1) \times n\).
\(2 \times (1+2+3+4+5+6) = 7 \times 6\). [span_35](start_span)[span_35](end_span)
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Select the correct option:
If you want to find the sum of all natural numbers from 15 to 40 inclusive, what is the correct algebraic expression using the \(S_n\) notation? [span_36](start_span)[span_36](end_span)
Concept:
1
To find a specific chunk of consecutive numbers, take the total sum up to the highest number, and subtract the total sum of the numbers directly preceding your chunk. [span_37](start_span)[span_37](end_span)
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Hint:
1
You want the chunk: [15, 16, 17... 40].
2
You do NOT want the chunk: [1, 2, 3... 14].
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Mistakes to Avoid:
!
Subtracting the lower bound: If you use \(S_{40} - S_{15}\), you are actively throwing the number 15 into the garbage. Your sum will only represent 16 through 40.
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Solution:
As taught in the textbook example for \(25 \dots 58\): [span_38](start_span)[span_38](end_span)
Total wanted: 15 to 40.
The unwanted segment is 1 to 14 (\(S_{14}\)).
Therefore, the equation is \(S_{40} - S_{14}\).
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