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Sequences and Series - Sum to n terms of Special Series

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of Special Series: Special series are those that do not directly follow the standard patterns of Arithmetic Progressions (AP) or Geometric Progressions (GP). To find their sum, we typically identify the general term ana_n and then apply the summation operator ∑\sum across its components.

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The Summation Operator (∑\sum): The symbol Σ\Sigma (Sigma) denotes the addition of a sequence of terms. Visually, if you imagine a stack of terms a1,a2,…,ana_1, a_2, \dots, a_n arranged vertically, the summation represents the total height of that stack. The expression ∑k=1nak\sum_{k=1}^{n} a_k tells us to start at k=1k=1 and add every term up to nn.

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Linearly of Summation: This property allows us to split a complex general term into simpler parts. If an=bn+cna_n = b_n + c_n, then ∑an=∑bn+∑cn\sum a_n = \sum b_n + \sum c_n. Additionally, any constant kk can be taken outside the summation: ∑k⋅an=k∑an\sum k \cdot a_n = k \sum a_n. This is like decomposing a complex shape into basic rectangles and triangles to find the total area.

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Sum of First nn Natural Numbers: This is the sum 1+2+3+⋯+n1 + 2 + 3 + \dots + n. Visually, this can be represented as a triangular arrangement of dots (triangular numbers). The total number of dots forms a shape whose area is exactly half of a rectangle with sides nn and n+1n+1.

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Sum of Squares of First nn Natural Numbers: This represents the sum 12+22+32+⋯+n21^2 + 2^2 + 3^2 + \dots + n^2. It grows much faster than the linear sum and is used when the general term of a series involves a quadratic expression n2n^2.

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Sum of Cubes of First nn Natural Numbers: This is the sum 13+23+33+⋯+n31^3 + 2^3 + 3^3 + \dots + n^3. A remarkable visual and algebraic property is that this sum is exactly equal to the square of the sum of the first nn natural numbers. If you visualize a square with side length ∑n\sum n, its total area equals the sum of these cubes.

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General Term Identification: The core strategy for solving special series involves finding the nthn^{th} term (ana_n) by observing the pattern of the factors. For example, in the series 1⋅2+2⋅3+3⋅41 \cdot 2 + 2 \cdot 3 + 3 \cdot 4, the nthn^{th} term is clearly n(n+1)n(n+1), which can be expanded to n2+nn^2 + n for easy summation.

📐Formulae

Sum of first nn natural numbers: ∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}

Sum of squares of first nn natural numbers: ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

Sum of cubes of first nn natural numbers: ∑k=1nk3=[n(n+1)2]2\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2

Sum of a constant cc up to nn terms: ∑k=1nc=nc\sum_{k=1}^{n} c = nc

General formula for sum: Sn=∑k=1nakS_n = \sum_{k=1}^{n} a_k

💡Examples

Problem 1:

Find the sum to nn terms of the series: 1⋅2+2⋅3+3⋅4+4⋅5+…1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + \dots

Solution:

  1. Identify the general term ana_n. The first factors are 1,2,3,4,…1, 2, 3, 4, \dots which is nn. The second factors are 2,3,4,5,…2, 3, 4, 5, \dots which is (n+1)(n+1).
  2. Thus, an=n(n+1)=n2+na_n = n(n+1) = n^2 + n.
  3. Apply the summation: Sn=∑k=1nak=∑k=1n(k2+k)S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^2 + k).
  4. Use linearity: Sn=∑k2+∑kS_n = \sum k^2 + \sum k.
  5. Substitute standard formulas: Sn=n(n+1)(2n+1)6+n(n+1)2S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}.
  6. Factor out n(n+1)2\frac{n(n+1)}{2}: Sn=n(n+1)2[2n+13+1]=n(n+1)2[2n+1+33]=n(n+1)(2n+4)6S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right] = \frac{n(n+1)}{2} \left[ \frac{2n+1+3}{3} \right] = \frac{n(n+1)(2n+4)}{6}.
  7. Simplify: Sn=n(n+1)2(n+2)6=n(n+1)(n+2)3S_n = \frac{n(n+1)2(n+2)}{6} = \frac{n(n+1)(n+2)}{3}.

Explanation:

To solve this, we first determine the pattern of the nthn^{th} term by looking at the individual components of each product. Once we have a polynomial expression for ana_n, we distribute the summation and use the standard power sum formulas.

Problem 2:

Find the sum to nn terms of the series whose nthn^{th} term is n2(n+3)n^2(n+3).

Solution:

  1. Expand the general term: an=n2(n+3)=n3+3n2a_n = n^2(n+3) = n^3 + 3n^2.
  2. Set up the sum: Sn=∑k=1n(k3+3k2)S_n = \sum_{k=1}^{n} (k^3 + 3k^2).
  3. Use linearity: Sn=∑k3+3∑k2S_n = \sum k^3 + 3\sum k^2.
  4. Substitute formulas: Sn=[n(n+1)2]2+3[n(n+1)(2n+1)6]S_n = \left[ \frac{n(n+1)}{2} \right]^2 + 3 \left[ \frac{n(n+1)(2n+1)}{6} \right].
  5. Simplify the second term: Sn=n2(n+1)24+n(n+1)(2n+1)2S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2}.
  6. Factor out n(n+1)2\frac{n(n+1)}{2}: Sn=n(n+1)2[n(n+1)2+(2n+1)]S_n = \frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + (2n+1) \right].
  7. Simplify inside the bracket: Sn=n(n+1)2[n2+n+4n+22]=n(n+1)(n2+5n+2)4S_n = \frac{n(n+1)}{2} \left[ \frac{n^2 + n + 4n + 2}{2} \right] = \frac{n(n+1)(n^2 + 5n + 2)}{4}.

Explanation:

This problem provides the general term directly. The strategy is to expand the polynomial, apply the summation to each power of nn, and factorize the resulting expression to get a clean final formula.