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.

All Sequence & Series notes NDA Maths strategy

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.

Concept 1 of 4

Sums of powers of natural numbers

Intuition

Three formulas sum the first

n

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 $n$ natural numbers:

$\displaystyle\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ .
$\displaystyle\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ .
$\displaystyle\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 = \left(\sum_{k=1}^{n} k\right)^2$ .

The three power sums

\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

\displaystyle\sum_{k=1}^{10} k^2

Use $\sum k^2 = \dfrac{n(n+1)(2n+1)}{6}$ with $n = 10$ .
$= \dfrac{10 \times 11 \times 21}{6} = \dfrac{2310}{6}$ .

Answer:

385

Practice this conceptself-check · 4 quick reps

Don't confuse the three power-sum formulas

Keep them distinct:

\sum k = \dfrac{n(n+1)}{2}

\sum k^2 = \dfrac{n(n+1)(2n+1)}{6}

, and

\sum k^3 = \left[\dfrac{n(n+1)}{2}\right]^2

. Only the cube-sum is a square. A frequent error is using the

\sum k

formula where

\sum k^2

is needed — the squares formula carries the extra

(2n+1)

and divides by 6. For

n = 4

\sum k^2 = \dfrac{4\cdot 5\cdot 9}{6} = 30

, not

\dfrac{4\cdot 5}{2} = 10

Concept 2 of 4

Arithmetic-geometric series (the S − rS trick)

Intuition

An arithmetic-geometric series has each term made of an AP factor times a GP factor —

1\cdot r + 2\cdot r^2 + 3\cdot r^3 + \cdots

. You sum it the same way you derived the GP sum: write

S

, write

rS

shifted one place, and subtract. The subtraction turns the AP coefficients into a constant, leaving an ordinary GP to sum.

Definition

For $S = \sum_{k=1}^{n} k\,r^k$ , form $rS$ , align by powers of $r$ , and subtract: $S - rS = (r + r^2 + \cdots + r^n) - n\,r^{n+1}$ . The bracket is a plain GP, so

S(1 - r) = \frac{r(1 - r^n)}{1 - r} - n\,r^{n+1}.

Solve for

S

. The same shift-and-subtract works for any AP

\times

GP term.

Worked example

Find

S = \displaystyle\sum_{k=1}^{n} k\,2^k = 1\cdot 2 + 2\cdot 2^2 + 3\cdot 2^3 + \cdots + n\cdot 2^n

Write $S = 1\cdot 2 + 2\cdot 2^2 + \cdots + n\cdot 2^n$ .
Write $2S = 1\cdot 2^2 + 2\cdot 2^3 + \cdots + n\cdot 2^{n+1}$ (every term shifted up one power).
Subtract: $S - 2S = (2 + 2^2 + \cdots + 2^n) - n\cdot 2^{n+1}$ .
The GP sums to $2^{n+1} - 2$ , so $-S = 2^{n+1} - 2 - n\cdot 2^{n+1}$ .
Thus $S = (n-1)\,2^{n+1} + 2$ .

Answer:

S = (n-1)\,2^{n+1} + 2

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Sequence & SeriesHARD

1\cdot 3 + 2\cdot 3^2 + 3\cdot 3^3 + \ldots + n\cdot 3^n = \dfrac{(2n-1)3^a + b}{4}

, then

a

and

b

are respectively

[Q28 · Sep · 2017]

Concept 3 of 4

Factorial sums — telescoping and remainders

Intuition

Factorials grow explosively, and two NDA shapes exploit that. A sum of

n\cdot n!

telescopes because

n\cdot n! = (n+1)! - n!

, so almost everything cancels. And a sum like

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: $n\cdot n! = (n+1)! - n!$ , so $\sum_{k=1}^{n} k\cdot k! = (n+1)! - 1$ . Remainder shape: to find $\big(\sum k!\big) \bmod m$ , note that $k! \equiv 0 \pmod{m}$ for all $k$ large enough that $m \mid k!$ — so only the small- $k$ terms contribute to the remainder.

Factorial telescoping

n\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! + \cdots + 100!

is divided by 12.

From $4!$ onward every factorial contains the factors $4 \times 3 = 12$ , so $4!, 5!, \ldots, 100!$ are all divisible by 12.
Only $1! + 2! + 3!$ can leave a remainder: $1 + 2 + 6 = 9$ .
Since $9 < 12$ , the remainder is $9$ .

Answer:

9

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Sequence & SeriesMODERATE

a_{n}=n(n!)

, then what is

a_{1}+a_{2}+a_{3}+\cdots+a_{10}

equal to?

[Q7 · Sep · 2021]

Drill 2 more on factorial sums — telescoping and remainders

Concept 4 of 4

Telescoping sums, repunits, and divisibility patterns

Intuition

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

\tfrac{10^n - 1}{9}

) and divisibility questions about

a^n \pm b^n

(which factor through

a \pm b

Definition

Telescoping: if $t_k = f(k) - f(k+1)$ , then $\sum_{k=1}^{n} t_k = f(1) - f(n+1)$ . A standard case: $\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1}$ , so the sum is $1 - \dfrac{1}{n+1} = \dfrac{n}{n+1}$ . Repunit: $\underbrace{11\ldots1}_{n} = \dfrac{10^n - 1}{9}$ . Factor identities: $a^n - b^n$ is divisible by $a - b$ (all $n$ ); $a^n + b^n$ is divisible by $a + b$ for odd $n$ .

Telescoping standard sum

\sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}

Worked example

Find

\dfrac{1}{1\cdot 2} + \dfrac{1}{2\cdot 3} + \dfrac{1}{3\cdot 4} + \cdots + \dfrac{1}{n(n+1)}

Split each term: $\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1}$ .
The sum becomes $\left(1 - \tfrac12\right) + \left(\tfrac12 - \tfrac13\right) + \cdots + \left(\tfrac1n - \tfrac{1}{n+1}\right)$ .
All the inner terms cancel, leaving $1 - \dfrac{1}{n+1}$ .

Answer:

\dfrac{n}{n+1}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Sequence & SeriesHARD

p = (1111\cdots\text{ up to }n\text{ digits})

, then what is the value of

9p^{2}+p

[Q5 · Sep · 2021]

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
$\sum k = \frac{n(n+1)}{2},\quad \sum k^2 = \frac{n(n+1)(2n+1)}{6},\quad \sum k^3 = \left[\frac{n(n+1)}{2}\right]^2$
Factorial sums — telescoping and remainders
Factorial telescoping
$n\cdot n! = (n+1)! - n! \;\Rightarrow\; \sum_{k=1}^{n} k\cdot k! = (n+1)! - 1$
Telescoping sums, repunits, and divisibility patterns
Telescoping standard sum
$\sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}$

Watch out for (1)

Don't confuse the three power-sum formulas
→ Sums of powers of natural numbers

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