Problem 268
Question
Consider \(n\) A.P.s whose first terms are \(1,2,3, \ldots \ldots \ldots, n\) and their common differences are \(1,2,3, \ldots \ldots \ldots .\) \(n\) respectively. If \(S_{i, i}\) denotes sum of \(i\) terms of \(i^{\text {th }}\) A.P., then find \(S_{1,1}+S_{2,2}+S_{3,3}+\cdots \cdots+S_{n, n} \cdot\\{\)
Step-by-Step Solution
Verified Answer
The sum \(S_{1,1}+S_{2,2}+S_{3,3}+ \ldots \ldots +S_{n,n}\) is given by \((n(n+1)/2)^2\).
1Step 1: Understanding the formula for the Sum of an Arithmetic Progression
The sum \(S_n\) of the first 'n' terms of an arithmetic progression where 'a' is the first term and 'd' is the common difference can be calculated as \(S_n = n/2 \[2a + (n-1)d\]\). This standard formula will be used to solve this problem. The 'nth' term in an arithmetic progression can be found by the formula \(T_n = a + (n-1)d\).
2Step 2: Applying the formula to our problem
In our exercise the first term 'a' and the common difference 'd' of each A.P are equal to 'i' (i.e., 'a' = 'd' = 'i'), where 'i' is the number of the A.P. sequence. Hence, the sum \(S_{i,i}\) of an 'i' terms of the 'i'-th A.P. can be evaluated as \(S_{i,i} = i/2 \[2i + (i-1)i\]\), which simplifies to \(S_{i,i} = i^3\). Hence, the required sum \(S_{1,1}+S_{2,2}+S_{3,3}+ \ldots \ldots +S_{n,n}\) becomes \(1^3 + 2^3 + 3^3 + \ldots \ldots + n^3\).
3Step 3: Simplifying the sum
The sum of cubes of the first 'n' natural numbers is given by the formula \((1^3 + 2^3 + 3^3 + \ldots \ldots + n^3) = (1 + 2 + 3 + \ldots + n)^2 = (n(n+1)/2)^2\).
Key Concepts
Sum of Arithmetic ProgressionAP Formula ApplicationSimplifying Series Sum
Sum of Arithmetic Progression
Understanding the sum of an arithmetic progression (AP) is essential for many mathematical applications, especially in arithmetical problems and sequence analysis found in competitive exams like the IIT JEE. An arithmetic progression is a sequence of numbers in which each term after the first is obtained by adding a constant, known as the common difference, to the previous term.
The sum of the first 'n' terms of an AP can be succinctly expressed using the formula: \[ S_n = \frac{n}{2}(2a + (n - 1)d) \], where \( S_n \) represents the sum of the 'n' terms, \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms. To simplify the calculation further, if all terms are needed, one might as well use the formula: \[ S_n = \frac{n}{2}(a + l) \], where \( l \) symbolizes the last term of the series.
Another useful aspect to remember is that the arithmetic mean of the first and the last term of an AP gives the average value of the series, which also aligns with the formula for the sum, reinforcing the conceptual integrity of these mathematical expressions. Through repetitive practice and intuitive understanding, solving for the sum of an arithmetic sequence can become a second nature.
The sum of the first 'n' terms of an AP can be succinctly expressed using the formula: \[ S_n = \frac{n}{2}(2a + (n - 1)d) \], where \( S_n \) represents the sum of the 'n' terms, \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms. To simplify the calculation further, if all terms are needed, one might as well use the formula: \[ S_n = \frac{n}{2}(a + l) \], where \( l \) symbolizes the last term of the series.
Another useful aspect to remember is that the arithmetic mean of the first and the last term of an AP gives the average value of the series, which also aligns with the formula for the sum, reinforcing the conceptual integrity of these mathematical expressions. Through repetitive practice and intuitive understanding, solving for the sum of an arithmetic sequence can become a second nature.
AP Formula Application
Application of the AP sum formula becomes interesting and less monotonous when the variables involved exhibit a pattern themselves, much like in the given IIT JEE problem. Here, we observe a grouping of multiple arithmetic progressions where the first term and the common difference for each AP is the same and corresponds to the position of the sequence in the grouping.
In such cases, we must adapt the standard formula to each individual AP by considering the first term (\( a \)), and the common difference (\( d \)) as 'i' for the \( i^{th} \) term. Thus, we have \[ S_{i,i} = \frac{i}{2} [2i + (i - 1)i] \], which then simplifies to \[ S_{i,i} = i^3 \].
This adaptation showcases that understanding the underlying principles of arithmetic progressions allows you to flexibly manipulate and apply formulas to a wide variety of problems, a skill highly valued in competitive examinations such as IIT JEE. The formula must be seen as a tool, adaptable to the specifics of the problem at hand, transforming a seemingly complex sequence into a straightforward solution.
In such cases, we must adapt the standard formula to each individual AP by considering the first term (\( a \)), and the common difference (\( d \)) as 'i' for the \( i^{th} \) term. Thus, we have \[ S_{i,i} = \frac{i}{2} [2i + (i - 1)i] \], which then simplifies to \[ S_{i,i} = i^3 \].
This adaptation showcases that understanding the underlying principles of arithmetic progressions allows you to flexibly manipulate and apply formulas to a wide variety of problems, a skill highly valued in competitive examinations such as IIT JEE. The formula must be seen as a tool, adaptable to the specifics of the problem at hand, transforming a seemingly complex sequence into a straightforward solution.
Simplifying Series Sum
Breaking down the sum of a series, especially when it involves a pattern of powers or multiples, is a critical skill in simplifying complex expressions. Recognizing that a series is comprised of individual terms that follow a certain rule is key to finding a general solution.
For instance, the sum of the cubes of the first 'n' natural numbers can be expressed as a square of the sum of 'n' natural numbers, which itself is an arithmetic progression. Thus, the resulting formula \[ (1^3 + 2^3 + 3^3 + \.\.\. + n^3) = \left( \frac{n(n+1)}{2} \right)^2 \] emerges.
Such simplifications are not only aesthetically pleasing but are immensely practical — they convert long, tedious calculations into manageable ones. They also illuminate deeper mathematical relationships, such as the power of squaring a simple AP sum to arrive at the summation of cubes. When approaching series sum problems, always look out for such patterns and formulas that condense the mathematical workload, thus making the problem approachable and solvable within the time constraints of a competitive exam setup.
For instance, the sum of the cubes of the first 'n' natural numbers can be expressed as a square of the sum of 'n' natural numbers, which itself is an arithmetic progression. Thus, the resulting formula \[ (1^3 + 2^3 + 3^3 + \.\.\. + n^3) = \left( \frac{n(n+1)}{2} \right)^2 \] emerges.
Such simplifications are not only aesthetically pleasing but are immensely practical — they convert long, tedious calculations into manageable ones. They also illuminate deeper mathematical relationships, such as the power of squaring a simple AP sum to arrive at the summation of cubes. When approaching series sum problems, always look out for such patterns and formulas that condense the mathematical workload, thus making the problem approachable and solvable within the time constraints of a competitive exam setup.
Other exercises in this chapter
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