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

Try it yourself

Find

\displaystyle\sum_{k=1}^{5} k^3

and check it equals

(\sum k)^2

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\sum_{k=1}^{n} k$ equals?
2.
$\sum_{k=1}^{20} k$ ?
3.
$\sum_{k=1}^{4} k^2$ ?
4.
$\sum k^3$ equals the square of?

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

Try it yourself

Evaluate

\displaystyle\sum_{k=1}^{3} k\,4^k = 1\cdot 4 + 2\cdot 4^2 + 3\cdot 4^3

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
In an arithmetic-geometric term $k\,r^k$ , which factor is the AP part?
2.
The summing trick is to subtract what from $S$ ?
3.
After subtracting, what kind of series is left?
4.
$\sum_{k=1}^{2} k\,3^k$ ?

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

Try it yourself

Use the telescoping identity to find

\displaystyle\sum_{k=1}^{5} k\cdot k!

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$n\cdot n!$ equals which difference?
2.
$\sum_{k=1}^{n} k\cdot k!$ equals?
3.
For $k \ge 4$ , is $k!$ divisible by 12?
4.
Remainder of $1! + 2! + 3! + 4!$ divided by 8?

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

Try it yourself

Show that

7^3 + 5^3

is divisible by 12, and give the quotient.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\dfrac{1}{k(k+1)}$ splits into?
2.
A repunit of $n$ ones equals?
3.
$a^n - b^n$ is always divisible by?
4.
$\sum_{k=1}^{3} \tfrac{1}{k(k+1)}$ ?

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

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Sequence & SeriesEASY

1! + 3! + 5! + 7! + \cdots + 199!

is divided by 24, what is the remainder?

[Q2 · Sep · 2023]

Example 2Sequence & SeriesMODERATE

What is the greatest integer among the following by which

5^5 + 7^5

is divisible?

[Q22 · Apr · 2018]

Example 3Sequence & SeriesHARD

What is the value of

2(2\times1)+3(3\times2\times1)+4(4\times3\times2\times1)+5(5\times4\times3\times2\times1)+\ldots+9(9\times8\times7\times6\times5\times4\times3\times2\times1)+2

[Q22 · Apr · 2022]

Example 4Sequence & SeriesEASY

Consider the following statements: 1. 2 + 4 + 6 + … + 2n = n² + n 2. The expression n² + n + 41 always gives a prime number for every natural number n

Which of the above statements is/are correct?

[Q10 · Sep · 2022]

Example 5Sequence & SeriesMODERATE

n \in N

, then

121^{n} - 25^{n} + 1900^{n} - (-4)^{n}

is divisible by which one of the following?

[Q1 · Apr · 2018]

Drill every past-year question on this subtopic

8 questions from the bank — paginated, with cart and Word-export support.

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Drill every past-year question on this subtopic

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