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NDA Maths · Sequence & Series

Special Series and Special Sums

The summation toolkit beyond AP and GP — power sums, arithmetic-geometric series, factorial sums, and telescoping — plus the number-pattern questions that ride on them.

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Why this matters

Eight PYQs, leaning HARD — this is where the difficulty of the chapter concentrates. The shapes are distinctive and each has a signature move: power sums use standard formulas, arithmetic-geometric sums use the subtract-r-times-the-sum trick, factorial sums telescope, and repunit / divisibility questions hinge on a clean factor identity. Recognise the shape, apply the move.

Test yourself — a quick 12-question Sequence & Series recall quiz, instant score.

Concept 1 of 4: Sums of powers of natural numbers

Three formulas sum the first nn natural numbers, their squares, and their cubes. They are worth memorising outright — they appear as building blocks inside many other series, and the cube-sum is just the square of the linear sum (a pleasing fact worth remembering).

Definition

For the first nn natural numbers:

  • ∑k=1nk=n(n+1)2\displaystyle\sum_{k=1}^{n} k = \frac{n(n+1)}{2}.
  • ∑k=1nk2=n(n+1)(2n+1)6\displaystyle\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}.
  • ∑k=1nk3=[n(n+1)2]2=(∑k=1nk)2\displaystyle\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 = \left(\sum_{k=1}^{n} k\right)^2.

The three power sums

∑k=n(n+1)2,∑k2=n(n+1)(2n+1)6,∑k3=[n(n+1)2]2\sum k = \frac{n(n+1)}{2},\quad \sum k^2 = \frac{n(n+1)(2n+1)}{6},\quad \sum k^3 = \left[\frac{n(n+1)}{2}\right]^2

Worked example

Find ∑k=110k2\displaystyle\sum_{k=1}^{10} k^2.
Practice this conceptself-check · 4 quick reps

Don't confuse the three power-sum formulas

Keep them distinct: ∑k=n(n+1)2\sum k = \dfrac{n(n+1)}{2}, ∑k2=n(n+1)(2n+1)6\sum k^2 = \dfrac{n(n+1)(2n+1)}{6}, and ∑k3=[n(n+1)2]2\sum k^3 = \left[\dfrac{n(n+1)}{2}\right]^2. Only the cube-sum is a square. A frequent error is using the ∑k\sum k formula where ∑k2\sum k^2 is needed — the squares formula carries the extra (2n+1)(2n+1) and divides by 6. For n=4n = 4: ∑k2=4⋅5⋅96=30\sum k^2 = \dfrac{4\cdot 5\cdot 9}{6} = 30, not 4⋅52=10\dfrac{4\cdot 5}{2} = 10.

Concept 2 of 4: Arithmetic-geometric series (the S − rS trick)

An arithmetic-geometric series has each term made of an AP factor times a GP factor — 1⋅r+2⋅r2+3⋅r3+⋯1\cdot r + 2\cdot r^2 + 3\cdot r^3 + \cdots. You sum it the same way you derived the GP sum: write SS, write rSrS shifted one place, and subtract. The subtraction turns the AP coefficients into a constant, leaving an ordinary GP to sum.

Definition

For S=∑k=1nk rkS = \sum_{k=1}^{n} k\,r^k, form rSrS, align by powers of rr, and subtract: S−rS=(r+r2+⋯+rn)−n rn+1S - rS = (r + r^2 + \cdots + r^n) - n\,r^{n+1}. The bracket is a plain GP, so

S(1−r)=r(1−rn)1−r−n rn+1.S(1 - r) = \frac{r(1 - r^n)}{1 - r} - n\,r^{n+1}.
Solve for SS. The same shift-and-subtract works for any AP ×\times GP term.

Worked example

Find S=∑k=1nk 2k=1⋅2+2⋅22+3⋅23+⋯+n⋅2nS = \displaystyle\sum_{k=1}^{n} k\,2^k = 1\cdot 2 + 2\cdot 2^2 + 3\cdot 2^3 + \cdots + n\cdot 2^n.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2017 · Q28Hard

Example 2 · Sequence & Series · Special Series and Special Sums

If 1⋅3+2⋅32+3⋅33+…+n⋅3n=(2n−1)3a+b41\cdot 3 + 2\cdot 3^2 + 3\cdot 3^3 + \ldots + n\cdot 3^n = \dfrac{(2n-1)3^a + b}{4}, then aa and bb are respectively

Concept 3 of 4: Factorial sums — telescoping and remainders

Factorials grow explosively, and two NDA shapes exploit that. A sum of n⋅n!n\cdot n! telescopes because n⋅n!=(n+1)!−n!n\cdot n! = (n+1)! - n!, so almost everything cancels. And a sum like 1!+2!+3!+⋯1! + 2! + 3! + \cdots modulo a small number is easy, because from some point on every factorial is divisible by that number — only the first few terms survive.

Definition

Telescoping identity: n⋅n!=(n+1)!−n!n\cdot n! = (n+1)! - n!, so ∑k=1nk⋅k!=(n+1)!−1\sum_{k=1}^{n} k\cdot k! = (n+1)! - 1. Remainder shape: to find (∑k!) mod m\big(\sum k!\big) \bmod m, note that k!≡0(modm)k! \equiv 0 \pmod{m} for all kk large enough that m∣k!m \mid k! — so only the small-kk terms contribute to the remainder.

Factorial telescoping

n⋅n!=(n+1)!−n!  ⇒  ∑k=1nk⋅k!=(n+1)!−1n\cdot n! = (n+1)! - n! \;\Rightarrow\; \sum_{k=1}^{n} k\cdot k! = (n+1)! - 1

Worked example

Find the remainder when 1!+2!+3!+⋯+100!1! + 2! + 3! + \cdots + 100! is divided by 12.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2021 · Q7Moderate

Example 3 · Sequence & Series · Special Series and Special Sums

If an=n(n!)a_{n}=n(n!), then what is a1+a2+a3+⋯+a10a_{1}+a_{2}+a_{3}+\cdots+a_{10} equal to?
Drill 2 more on factorial sums — telescoping and remainders

Concept 4 of 4: Telescoping sums, repunits, and divisibility patterns

Telescoping is the most satisfying summation: rewrite each term as a difference so that consecutive pieces cancel, leaving only the first and last. The same "find the hidden structure" instinct cracks repunit problems (a string of 1s is 10n−19\tfrac{10^n - 1}{9}) and divisibility questions about an±bna^n \pm b^n (which factor through a±ba \pm b).

Definition

Telescoping: if tk=f(k)−f(k+1)t_k = f(k) - f(k+1), then ∑k=1ntk=f(1)−f(n+1)\sum_{k=1}^{n} t_k = f(1) - f(n+1). A standard case: 1k(k+1)=1k−1k+1\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1}, so the sum is 1−1n+1=nn+11 - \dfrac{1}{n+1} = \dfrac{n}{n+1}. Repunit: 11…1⏟n=10n−19\underbrace{11\ldots1}_{n} = \dfrac{10^n - 1}{9}. Factor identities: an−bna^n - b^n is divisible by a−ba - b (all nn); an+bna^n + b^n is divisible by a+ba + b for odd nn. Two number-pattern cautions the statement-checks are built on: (1) a formula that is prime for many consecutive nn need not be prime for all — Euler's n2+n+41n^2 + n + 41 is prime for n=0,1,…,39n = 0, 1, \dots, 39 and composite at n=40n = 40 (=412= 41^2) and n=41n = 41; "n2+n+41n^2+n+41 is always prime" is FALSE. (2) "Is 2m(m+2)2m(m+2) ever a perfect square?" — for a bounded search, note 2m(m+2)=2(m+1)2−22m(m+2) = 2(m+1)^2 - 2 sits strictly between (m+1)2(m+1)^2 and (m+2)2(m+2)^2 for m≥2m \ge 2 (check m=1m=1: 6; m=2m=2: 16 =42=4^2 — so exactly one mm works). Test small cases before trusting a pattern, and bracket the expression between consecutive squares to stop the search.

Telescoping standard sum

∑k=1n1k(k+1)=1−1n+1=nn+1\sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}

Worked example

Find 11⋅2+12⋅3+13⋅4+⋯+1n(n+1)\dfrac{1}{1\cdot 2} + \dfrac{1}{2\cdot 3} + \dfrac{1}{3\cdot 4} + \cdots + \dfrac{1}{n(n+1)}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2021 · Q5Hard

Example 4 · Sequence & Series · Special Series and Special Sums

If p=(1111⋯ up to n digits)p = (1111\cdots\text{ up to }n\text{ digits}), then what is the value of 9p2+p9p^{2}+p?
Drill 3 more on telescoping sums, repunits, and divisibility patterns

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Sums of powers of natural numbers

    The three power sums

    ∑k=n(n+1)2,∑k2=n(n+1)(2n+1)6,∑k3=[n(n+1)2]2\sum k = \frac{n(n+1)}{2},\quad \sum k^2 = \frac{n(n+1)(2n+1)}{6},\quad \sum k^3 = \left[\frac{n(n+1)}{2}\right]^2
  • Factorial sums — telescoping and remainders

    Factorial telescoping

    n⋅n!=(n+1)!−n!  ⇒  ∑k=1nk⋅k!=(n+1)!−1n\cdot n! = (n+1)! - n! \;\Rightarrow\; \sum_{k=1}^{n} k\cdot k! = (n+1)! - 1
  • Telescoping sums, repunits, and divisibility patterns

    Telescoping standard sum

    ∑k=1n1k(k+1)=1−1n+1=nn+1\sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}

Watch out for (1)

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