Mastering Sequences and Series Ex 9.4: Sum of Special Series

Understanding Exercise 9.4: The Sum of Special Series

Exercise 9.4 in your NCERT Class 11 Mathematics textbook for Sequences and Series primarily deals with finding the sum of series whose general terms (nth terms) are not directly an AP or a GP. Instead, the nth term often involves products of terms from an AP, or it might be a polynomial in 'n'. The core strategy involves two main steps:

1. Finding the General Term (nth term): The first and often most challenging step is to identify a pattern and express the nth term of the given series, denoted as $a_n$ or $T_n$, as a function of 'n'. This might require careful observation of the terms' structure or applying the 'method of differences' when the terms' differences form an AP or GP.
2. Summation Using Standard Formulas: Once the nth term, $T_n$, is found, the sum of the first 'n' terms, denoted as $S_n$, is calculated by summing $T_n$ from $k=1$ to $n$, i.e., $S_n = \sum_{k=1}^{n} T_k$. This step often requires the application of standard summation formulas for the sum of the first 'n' natural numbers, sum of squares of the first 'n' natural numbers, and sum of cubes of the first 'n' natural numbers. Mastering these formulas and their algebraic manipulation is key to success in this exercise.

Essential Summation Formulas and Concepts

Sum of First n Natural Numbers

The sum of the first 'n' positive integers is given by the formula: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$. This is crucial for series where the general term has a linear 'n' component.

Sum of Squares of First n Natural Numbers

The sum of the squares of the first 'n' positive integers is given by: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$. This formula is used when the general term involves a quadratic 'n' component, like $n^2$ or $n(n+1)$.

Sum of Cubes of First n Natural Numbers

The sum of the cubes of the first 'n' positive integers is given by: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2$. This formula becomes relevant when the general term contains a cubic 'n' component, such as $n^3$ or $n^2(n+1)$.

General Term ($T_n$ or $a_n$)

The general term, or nth term, of a sequence or series is an algebraic expression that describes any term in the sequence based on its position 'n'. For example, if a series is $1, 4, 9, 16, \ldots$, its general term is $T_n = n^2$.

Finding the General Term using the Method of Differences

1. Step 1: Write down the given series and its terms — Clearly list the first few terms of the series, for example, $T_1, T_2, T_3, \ldots$. This helps in observing patterns and differences.
2. Step 2: Calculate the differences between consecutive terms — Find the first order differences: $D_1 = T_2 - T_1, D_2 = T_3 - T_2, \ldots$. If these differences form an AP or GP, you might be able to find the nth term of these differences. If not, proceed to the next step.
3. Step 3: Calculate the differences of the differences (second order differences) — If the first order differences didn't form a simple progression, find the differences between the first order differences: $D_1' = D_2 - D_1, D_2' = D_3 - D_2, \ldots$. Continue this process until you find a constant difference or a progression that is easily identifiable (e.g., an AP or GP).
4. Step 4: Determine the form of the nth term — If the first differences are constant, the nth term $T_n$ is a linear expression in 'n' (an AP). If the second differences are constant, $T_n$ is a quadratic expression in 'n' (e.g., $An^2+Bn+C$). If the third differences are constant, $T_n$ is a cubic expression, and so on. If the differences form a GP, $T_n$ involves exponential terms. In general, if the $k^{th}$ order differences are constant, $T_n$ is a polynomial of degree $k$ in 'n'.
5. Step 5: Set up equations to find coefficients — Assume the form of $T_n$ (e.g., $An+B$, $An^2+Bn+C$, etc.). Substitute the first few terms of the series (e.g., $n=1, n=2, n=3$) into the assumed form of $T_n$ to create a system of linear equations. Solve these equations to find the values of the coefficients (A, B, C, etc.).
6. Step 6: Verify the nth term and find the sum — Once $T_n$ is found, substitute it into the general summation formula: $S_n = \sum_{k=1}^{n} T_k$. Then, apply the standard summation formulas for $\sum k, \sum k^2, \sum k^3$ as needed, and simplify the expression to get the sum of the series.

Worked Examples for Sequences and Series Ex 9.4

• Example 1: Find the sum to n terms of the series $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots$ Step 1: Find the general term ($T_n$) Observe the pattern: The first part of each term is $1, 2, 3, \ldots$, which is 'n'. The second part of each term is $2, 3, 4, \ldots$, which is $(n+1)$. So, the general term is $T_n = n(n+1) = n^2 + n$. Step 2: Find the sum ($S_n$) $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k)$ Using the properties of summation, $S_n = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$ Step 3: Apply the standard summation formulas We know: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ Substitute these into the expression for $S_n$: $S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$ Step 4: Simplify the expression Take common factors $ \frac{n(n+1)}{2} $: $S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right]$ $S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1+3}{3} \right]$ $S_n = \frac{n(n+1)}{2} \left[ \frac{2n+4}{3} \right]$ $S_n = \frac{n(n+1)}{2} \cdot \frac{2(n+2)}{3}$ $S_n = \frac{n(n+1)(n+2)}{3}$ Final Answer: The sum to n terms of the series is $\frac{n(n+1)(n+2)}{3}$.
• Example 2: Find the sum to n terms of the series whose nth term is $n(n+1)(n+4)$ Step 1: The general term ($T_n$) is already given $T_n = n(n+1)(n+4)$ Expand the expression: $T_n = (n^2+n)(n+4) = n^3 + 4n^2 + n^2 + 4n = n^3 + 5n^2 + 4n$ Step 2: Find the sum ($S_n$) $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^3 + 5k^2 + 4k)$ Using properties of summation: $S_n = \sum_{k=1}^{n} k^3 + 5 \sum_{k=1}^{n} k^2 + 4 \sum_{k=1}^{n} k$ Step 3: Apply the standard summation formulas Substitute the formulas: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2$ $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ So, $S_n = \left[ \frac{n(n+1)}{2} \right]^2 + 5 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 4 \left[ \frac{n(n+1)}{2} \right]$ Step 4: Simplify the expression $S_n = \frac{n^2(n+1)^2}{4} + \frac{5n(n+1)(2n+1)}{6} + \frac{4n(n+1)}{2}$ Factor out $ \frac{n(n+1)}{2} $: $S_n = \frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + \frac{5(2n+1)}{3} + 4 \right]$ Find a common denominator (6) inside the bracket: $S_n = \frac{n(n+1)}{2} \left[ \frac{3n(n+1) + 10(2n+1) + 24}{6} \right]$ $S_n = \frac{n(n+1)}{12} [3n^2 + 3n + 20n + 10 + 24]$ $S_n = \frac{n(n+1)}{12} [3n^2 + 23n + 34]$ Final Answer: The sum to n terms is $\frac{n(n+1)(3n^2 + 23n + 34)}{12}$.

Exam Tips for Sequences and Series Ex 9.4

To excel in problems from Exercise 9.4, remember these crucial tips:

• Memorize Formulas: The summation formulas for $\sum k, \sum k^2,$ and $\sum k^3$ are fundamental. Practice writing them down until they are second nature.
• Correctly Identify the nth Term: This is the most critical step. Spend sufficient time observing patterns. For series like $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots$, identify the pattern for each factor in the term. If direct observation is hard, use the method of differences carefully.
• Algebraic Simplification: After applying summation formulas, the resulting expression can be complex. Practice careful algebraic manipulation, especially factoring out common terms like $n(n+1)$ or $n(n+1)/2$, to simplify the final answer. This saves time and reduces calculation errors.
• Check Your Work: For simpler series, calculate the sum for a small 'n' (e.g., n=1 or n=2) using the derived formula and compare it to the direct sum of the first few terms. This helps catch errors in your $T_n$ or $S_n$ derivation.

Practice Questions with Solutions

• Q: Find the sum to n terms of the series $1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots$ A: Step 1: Find the general term ($T_n$). Observe the first factors: $1, 3, 5, \ldots$ This is an AP with $a=1, d=2$. So, its nth term is $1 + (n-1)2 = 2n-1$. Observe the second factors: $3, 5, 7, \ldots$ This is an AP with $a=3, d=2$. So, its nth term is $3 + (n-1)2 = 2n+1$. Thus, $T_n = (2n-1)(2n+1) = 4n^2 - 1$. Step 2: Find the sum ($S_n$). $S_n = \sum_{k=1}^{n} (4k^2 - 1) = 4 \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} 1$ Step 3: Apply the standard summation formulas. $S_n = 4 \left[ \frac{n(n+1)(2n+1)}{6} \right] - n$ $S_n = \frac{2n(n+1)(2n+1)}{3} - n$ Step 4: Simplify the expression. $S_n = n \left[ \frac{2(n+1)(2n+1)}{3} - 1 \right]$ $S_n = n \left[ \frac{2(2n^2+3n+1) - 3}{3} \right]$ $S_n = n \left[ \frac{4n^2+6n+2-3}{3} \right]$ $S_n = \frac{n(4n^2+6n-1)}{3}$ Final answer: The sum to n terms is $\frac{n(4n^2+6n-1)}{3}$.
• Q: Find the sum to n terms of the series $1^2 + (1^2+2^2) + (1^2+2^2+3^2) + \ldots$ A: Step 1: Find the general term ($T_n$). The nth term of this series is the sum of the squares of the first 'n' natural numbers. $T_n = \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} = \frac{n(2n^2+3n+1)}{6} = \frac{2n^3+3n^2+n}{6}$ Step 2: Find the sum ($S_n$). $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{2k^3+3k^2+k}{6} = \frac{1}{6} \left[ 2 \sum_{k=1}^{n} k^3 + 3 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \right]$ Step 3: Apply the standard summation formulas. $S_n = \frac{1}{6} \left[ 2 \left( \frac{n(n+1)}{2} \right)^2 + 3 \left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{n(n+1)}{2} \right]$ $S_n = \frac{1}{6} \left[ \frac{2n^2(n+1)^2}{4} + \frac{3n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right]$ $S_n = \frac{1}{6} \left[ \frac{n^2(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{2} + \frac{n(n+1)}{2} \right]$ Step 4: Simplify the expression. Factor out $ \frac{n(n+1)}{2} $ from the bracket: $S_n = \frac{1}{6} \cdot \frac{n(n+1)}{2} [n(n+1) + (2n+1) + 1]$ $S_n = \frac{n(n+1)}{12} [n^2+n + 2n+1 + 1]$ $S_n = \frac{n(n+1)}{12} [n^2+3n+2]$ $S_n = \frac{n(n+1)}{12} [(n+1)(n+2)]$ $S_n = \frac{n(n+1)^2(n+2)}{12}$ Final answer: The sum to n terms is $\frac{n(n+1)^2(n+2)}{12}$.
• Q: Find the sum to n terms of the series whose nth term is $2n^2 + 3n$. A: Step 1: The general term ($T_n$) is already given. $T_n = 2n^2 + 3n$ Step 2: Find the sum ($S_n$). $S_n = \sum_{k=1}^{n} (2k^2 + 3k) = 2 \sum_{k=1}^{n} k^2 + 3 \sum_{k=1}^{n} k$ Step 3: Apply the standard summation formulas. $S_n = 2 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 3 \left[ \frac{n(n+1)}{2} \right]$ Step 4: Simplify the expression. $S_n = \frac{n(n+1)(2n+1)}{3} + \frac{3n(n+1)}{2}$ Take common factors $ \frac{n(n+1)}{2} $: $S_n = \frac{n(n+1)}{2} \left[ \frac{2(2n+1)}{3} + 3 \right]$ $S_n = \frac{n(n+1)}{2} \left[ \frac{4n+2+9}{3} \right]$ $S_n = \frac{n(n+1)(4n+11)}{6}$ Final answer: The sum to n terms is $\frac{n(n+1)(4n+11)}{6}$.
• Q: Find the sum of the series $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \ldots$ up to n terms. A: Step 1: Find the general term ($T_n$). Observe the patterns for each factor: First factors: $1, 2, 3, \ldots \implies n$ Second factors: $2, 3, 4, \ldots \implies n+1$ Third factors: $3, 4, 5, \ldots \implies n+2$ So, $T_n = n(n+1)(n+2)$ Expand $T_n$: $T_n = (n^2+n)(n+2) = n^3 + 2n^2 + n^2 + 2n = n^3 + 3n^2 + 2n$ Step 2: Find the sum ($S_n$). $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^3 + 3k^2 + 2k)$ $S_n = \sum_{k=1}^{n} k^3 + 3 \sum_{k=1}^{n} k^2 + 2 \sum_{k=1}^{n} k$ Step 3: Apply the standard summation formulas. $S_n = \left[ \frac{n(n+1)}{2} \right]^2 + 3 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 2 \left[ \frac{n(n+1)}{2} \right]$ $S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1)$ Step 4: Simplify the expression. Factor out $ \frac{n(n+1)}{4} $ (since the first term has denominator 4): $S_n = \frac{n(n+1)}{4} \left[ n(n+1) + 2(2n+1) + 4 \right]$ $S_n = \frac{n(n+1)}{4} [n^2+n + 4n+2 + 4]$ $S_n = \frac{n(n+1)}{4} [n^2+5n+6]$ Factor the quadratic term $n^2+5n+6 = (n+2)(n+3)$ $S_n = \frac{n(n+1)(n+2)(n+3)}{4}$ Final answer: The sum to n terms is $\frac{n(n+1)(n+2)(n+3)}{4}$.

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