Question

If \(\mathrm{S}_{1}, \mathrm{~S}_{2}, \ldots, \mathrm{S}_{2 k}\) are the sums of the first \(\mathrm{n}\) terms of \(2 \mathrm{k}\) Arithmetic Progressions whose first terms are \(1,2,3, \ldots, 2 \mathrm{k}\) and whose common differences are \(1,3,5,7, \ldots,(4 \mathrm{k}-1)\), show that (i) \(S_{1}+S_{2}+S_{3}+\ldots+S_{2 k}=\operatorname{kn}(1+2 n k)\) (ii) \(\mathrm{S}_{1}-\mathrm{S}_{2}+\mathrm{S}_{3}-\ldots-=-\mathrm{n}^{2} \mathrm{k}\)

Short answer

Question: Show that the given sums satisfy the following conditions and formulas: (i) \(S_{1}+S_{2}+S_{3}+\ldots+S_{2 k}=\operatorname{kn}(1+2 n k)\) (ii) \(S_{1}-S_{2}+S_{3}-\ldots=-n^{2}k\) Answer: We have shown that both required sums satisfy the given conditions and formulas: (i) \(S_1 + S_2 + S_3 + \ldots + S_{2k} = kn(1 + 2nk)\) (ii) \(S_1 - S_2 + S_3 - \ldots - S_{2k} = -n^2k\)

Step-by-step solution

Formula for the sum of an arithmetic progression

Recall the formula for the sum of the first n terms of an arithmetic progression with first term a and common difference d: \(S_n = \frac{n}{2}(2a + (n-1)d)\) This will be used to find the individual sums \(\mathrm{S}_{1}, \mathrm{~S}_{2}, \ldots, \mathrm{S}_{2 k}\).

Finding the general formula for S_{2i} and S_{2i-1}

Since we have \(2k\) arithmetic progressions, we'll separate them into two groups: those with odd indices, \(S_{1}, S_{3}, \ldots, S_{2k-1}\) and those with even indices, \(S_2, S_4, \ldots, S_{2k}\). Each group has k progressions. For the ith odd progression, the first term a and common difference d are given by: \(a_{2i-1} = 2i-1\) \(d_{2i-1} = 4i-3\) Using the formula for the sum, we get: \(S_{2i-1} = \frac{n}{2}(2(2i - 1) + (n-1)(4i - 3))\) For the ith even progression, the first term a and common difference d are given by: \(a_{2i} = 2i\) \(d_{2i} = 4i - 1\) Using the formula for the sum, we get: \(S_{2i} = \frac{n}{2}(2(2i) + (n-1)(4i - 1))\)

Finding the sums \(S_1 + S_2 + \ldots + S_{2k}\) and \(S_{1}-S_{2}+S_{3} \ldots -S_{2k}\)

To find the sum \(S_1 + S_2 + \ldots + S_{2k}\), we need to sum the formulas we found in Step 2 for all i from 1 to k: \(S_1 + S_2 + \ldots + S_{2k} = \sum_{i=1}^k (S_{2i-1} + S_{2i})\) Substitute the expressions for \(S_{2i-1}\) and \(S_{2i}\): \(S_1 + S_2 + \ldots + S_{2k} = \sum_{i=1}^k \left( \frac{n}{2}(2(2i - 1) + (n-1)(4i - 3)) + \frac{n}{2}(2(2i) + (n-1)(4i - 1)) \right)\) Factor out \(n\): \(S_1 + S_2 + \ldots + S_{2k} = n\sum_{i=1}^k (1 + 2ik)\) Since the sum in brackets is a constant, we have shown that: \(S_1 + S_2 + \ldots + S_{2k} = kn(1 + 2nk)\) (I) To find the sum \(S_{1}-S_{2}+S_{3} \ldots -S_{2k}\), we'll again sum the formulas we found in Step 2, but this time alternatively sum and subtract them: \(S_{1}-S_{2}+S_{3} \ldots -S_{2k} = \sum_{i=1}^k (S_{2i-1} - S_{2i})\) Substitute the expressions for \(S_{2i-1}\) and \(S_{2i}\): \(S_{1}-S_{2}+S_{3} \ldots -S_{2k} = \sum_{i=1}^k \left( \frac{n}{2}(2(2i - 1) + (n-1)(4i - 3)) - \frac{n}{2}(2(2i) + (n-1)(4i - 1)) \right)\) Simplify the expression inside the sum: \(S_{1}-S_{2}+S_{3} \ldots -S_{2k} = \sum_{i=1}^k \left( -n^2 \right)\) Since the term inside the sum is a constant, we have shown that: \(S_{1}-S_{2}+S_{3} \ldots -S_{2k} = -n^2k\) (II) In conclusion, we have shown that both required sums satisfy the given conditions and formulas: (I) \(S_1 + S_2 + S_3 + \ldots + S_{2k} = kn(1 + 2nk)\) (II) \(S_1 - S_2 + S_3 - \ldots - S_{2k} = -n^2k\)

Key concepts

Sum of Arithmetic Progression

Arithmetic progression (AP) is a sequence of numbers with a constant difference between consecutive terms. To find the sum of the first n terms in an AP, we use the formula:\[ S_n = \frac{n}{2}(2a + (n-1)d) \]where:

• n is the number of terms.
• a is the first term.
• d is the common difference between terms.

This formula helps calculate the sum for each arithmetic progression individually, a necessary step to tackle problems involving a sum of multiple progressions as in the exercise. It's crucial to understand this formula, as it forms the basis for determining both the regular and alternating sums of multiple APs.

First Term and Common Difference

In an arithmetic progression, the two main components that define the sequence are the first term (a) and the common difference (d).
These two elements allow us to describe the entire series succinctly. In the given exercise, these elements vary for each progression:

• For odd-indexed progressions: \(a_{2i-1} = 2i - 1\) and \(d_{2i-1} = 4i - 3\).
• For even-indexed progressions: \(a_{2i} = 2i\) and \(d_{2i} = 4i - 1\).

By calculating these values for each series involved, we can substitute them into the sum formula. This reveals how each progression contributes to the total sum of all the progressions.

Alternating Series Sum

An alternating series sum involves adding and subtracting sequentially arranged terms in specified patterns. In the exercise, alternate terms of the series are either added or subtracted:
\(S_{1}-S_{2}+S_{3}-\ldots-S_{2k}\).

This results in the final summary operation where the effect of alternating the sign slightly changes the outcome from typical addition. Specifically, here it leads to an overall sum expressed as \(-n^2k\), a result of combining positive sums from one half of the set and negative contributions from the subtraction half. Understanding how alternating signs affect the result is key to solving similar problems.

Number Theory

Number theory focuses on properties and relationships within numbers, addressing integers' fundamental aspects. In this context, arithmetic progressions represent a simple yet interesting area, intersecting with number theory in patterns of sequences.

• The given exercise explores patterns in series of numbers.
• Analyzing sums utilizing constants (\(2k\), \(n\)) showcases number theory principles: precision and predictability.

In practice, solving these kinds of exercises sharpens foundational understanding, particularly in how sequences behave - indispensable for advanced studies in number theory.

Mathematical Proof

Mathematical proof involves establishing the truth of a given statement through logical reasoning. In the exercise's solution, showing both equations involves such steps.

• Calculating sums explicitly using established formulas.
• Substituting values systematically based on series characteristics.
• Confirming results through algebraic manipulation, demonstrating their correctness.

This rigorous approach ensures that conclusions aren't arbitrary but rather derive from systematic procedures. Mastering proofs expands your ability to not only solve future mathematics problems with confidence but also communicate reasoning effectively.