krit.club logo

Exploring some more Progressions - Method of Differences and Combinatorics-advanced

Grade 9CBSE

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

🔑Concepts

•

The Method of Differences is used to find the general term TnT_n and the sum SnS_n of a sequence when the differences between consecutive terms form an Arithmetic Progression (AP) or a Geometric Progression (GP).

•

If the kk-th order differences of a sequence become constant, the general term TnT_n is a polynomial in nn of degree kk. For example, if the first differences are in AP, the second differences are constant, and Tn=an2+bn+cT_n = an^2 + bn + c.

•

The sum of special series involves calculating ∑n\sum n, ∑n2\sum n^2, and ∑n3\sum n^3. These are often used as components when finding the sum of a sequence after determining its general term TnT_n.

•

Combinatorics in sequences involves using the properties of binomial coefficients, such as (nr)\binom{n}{r}, to simplify sums. A key identity is the Hockey-stick Identity: ∑i=rn(ir)=(n+1r+1)\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}.

•

The Telescoping Method (or Vn method) expresses the general term TrT_r as a difference of two consecutive terms of another sequence, Tr=Vr−Vr−1T_r = V_r - V_{r-1}. The sum is then ∑r=1nTr=Vn−V0\sum_{r=1}^{n} T_r = V_n - V_0.

📐Formulae

∑r=1nr=n(n+1)2\sum_{r=1}^{n} r = \frac{n(n+1)}{2}

∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}

∑r=1nr3=[n(n+1)2]2\sum_{r=1}^{n} r^3 = \left[ \frac{n(n+1)}{2} \right]^2

Tn=a(n−10)+Δ1(n−11)+Δ2(n−12)+…T_n = a \binom{n-1}{0} + \Delta_1 \binom{n-1}{1} + \Delta_2 \binom{n-1}{2} + \dots

∑r=1nr(r+1)(r+2)…(r+k−1)=n(n+1)(n+2)…(n+k)k+1\sum_{r=1}^{n} r(r+1)(r+2)\dots(r+k-1) = \frac{n(n+1)(n+2)\dots(n+k)}{k+1}

💡Examples

Problem 1:

Find the nn-th term and the sum of the first nn terms of the sequence: 3,7,13,21,31,…3, 7, 13, 21, 31, \dots

Solution:

Let the sequence be S=3+7+13+21+⋯+TnS = 3 + 7 + 13 + 21 + \dots + T_n The first differences are: 7−3=47-3=4, 13−7=613-7=6, 21−13=821-13=8, 31−21=1031-21=10. The differences 4,6,8,10,…4, 6, 8, 10, \dots form an AP with a=4a=4 and d=2d=2. Since the second difference is constant (6−4=26-4=2), TnT_n is a quadratic: Tn=an2+bn+cT_n = an^2 + bn + c. Using the formula Tn=T1(n−10)+Δ1(n−11)+Δ2(n−12)T_n = T_1 \binom{n-1}{0} + \Delta_1 \binom{n-1}{1} + \Delta_2 \binom{n-1}{2}: Tn=3(1)+4(n−1)+2(n−1)(n−2)2T_n = 3(1) + 4(n-1) + 2\frac{(n-1)(n-2)}{2} Tn=3+4n−4+n2−3n+2=n2+n+1T_n = 3 + 4n - 4 + n^2 - 3n + 2 = n^2 + n + 1 To find SnS_n: Sn=∑r=1n(r2+r+1)=∑r2+∑r+∑1S_n = \sum_{r=1}^{n} (r^2 + r + 1) = \sum r^2 + \sum r + \sum 1 Sn=n(n+1)(2n+1)6+n(n+1)2+nS_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n Sn=n6[(n+1)(2n+1)+3(n+1)+6]S_n = \frac{n}{6} [(n+1)(2n+1) + 3(n+1) + 6] Sn=n(2n2+6n+10)6=n(n2+3n+5)3S_n = \frac{n(2n^2 + 6n + 10)}{6} = \frac{n(n^2 + 3n + 5)}{3}

Explanation:

We identify that the first differences form an AP, implying the second differences are constant. We use the general form for the nn-th term of such a series and then apply summation formulas for n2n^2, nn, and constants.

Problem 2:

Evaluate the sum: S=11×2+12×3+13×4+⋯+1n(n+1)S = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{n(n+1)}

Solution:

The general term is Tr=1r(r+1)T_r = \frac{1}{r(r+1)}. Using partial fractions or the method of differences: Tr=(r+1)−rr(r+1)=1r−1r+1T_r = \frac{(r+1) - r}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1} Now, sum from r=1r=1 to nn: Sn=(11−12)+(12−13)+⋯+(1n−1n+1)S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \dots + \left( \frac{1}{n} - \frac{1}{n+1} \right) Most terms cancel out (telescoping): Sn=1−1n+1=n+1−1n+1=nn+1S_n = 1 - \frac{1}{n+1} = \frac{n+1-1}{n+1} = \frac{n}{n+1}

Explanation:

This is a telescoping series. By splitting each term into a difference of two fractions, all intermediate terms cancel out, leaving only the first and last parts.