Samacheer Class 11 Maths - Binomial Theorem, Sequences and Series

(i) 102 4
102 4 = (100 + 2) 4
= 100000000 + 8 × 1000000 + 24 × 10000 + 3200 + 16
= 100000000 + 8000000 + 240000 + 3216
= 108243216
(ii) 99 4
99 4 = (100 – 1) 4
= 100000000 – 4 × 1000000 + 6 × 10000 – 400 + 1
= 100000000 – 4000000 + 60000 -400+1
= 96059601
(iii) 9 7
9 7 = (10 – 1) 7
= 10 3 (10 4 – 7 × 10 3 + 21 × 10 2 – 35 × 10 + 35) – 21 × 100 + 70 – 1
= 10 3 (10000 – 7000 + 2100 – 350 + 35 ) – 2100 + 70 – 1
= 10 3 (12135 – 7350) – 2031
= 10 3 × 4785 – 2031
= 4785000 – 2031
= 4782969

General term T r+1 = nC r x n-r. a r
∴ In the expansion \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)
To find the coefficient of x 15, Put 20 – 5r = 15
20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 1
∴ T 1 + 1 = 10C 1 x 20 – 5 ⇒ T 2 = 10. x 15
∴ The coefficient of x 15 is 10

The general terms is T r+1 = nC r a n-r. a r
∴ The general term in the expansion of \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\) is
To find the coefficient of x 6, Put 12 – 5r = 6
12 – 6 = 5r ⇒ 5r = 6 ⇒ r = \(\frac{6}{5}\) which is impossible.
∴ There is no x 6 term in the expansion.
To find the coefficient of x 2,
Put 12 – 5r = 2
⇒ 12 – 2 = 5r
⇒ 10 = 5r
⇒ r = \(\frac{10}{5}\) = 5
Substituting in (1) we have
T 3 = 15
∴ The coefficient of x 2 is 15

= 10x 4 × 1 + 5 × 1225 x 4 + 19600 x 4 + 500 x 4
= x 4 (10 + 6125 + 19600 + 500) = 26235. x 4
∴ The coefficient of x 4 is 26235.

Consider 3 600
3 600 = (3 2 ) 300 = 9 300 = (1o – 1) 300
= 10 300 – 300 (10) 299 + ………………. + 300 C 1 × 10 × – 1 + 1 × 1 × 1
= 10 300 – 300 (1o) 299 + …………….. – 300 × 10 + 1
= 10 300 – 300 × 10 299 + ……………… – 3000 + 1
All the terms except the last are multiples of 100 and hence divisible by 100.
∴ The last two digits will be 01.

which is divisible by 64 for all positive integer n.
∴ 9 n – 8n – 1 is divisible by 64 for all positive integer n.
Put n = n + 1 we get
9 n + 1 – 8 (n + 1) – 1 is divisible by 64 for all possible integer n
(9 n + 1 – 8n – 8 – 1) is divisible by 64
∴ 9 n + 1 – 8n – 9 is always divisible by 64

Given (1 + x) n.
General term T r+1 = nC r x n – r. a r
∴ The general term in the expansion of (1 + x) n is T r+1 = nC r. (1) n – r. x r
T r+1 = nC r. x r ……….. (1)
∴ Coefficient of x r is nC r,
Put r = n – r in (1)
T n – r + 1 = nC n – r. x n-r
∴ The coefficient of x n-r is nC n-r …………. (2)
we know nC r = nC n-r
∴ The coefficient of x r and coefficient of x n-r are equal, (by (1) & (2))

In (a + b) n general term is t r + 1 = n C r a n – r b r
So, t 4 = t 3 + 1 = n C 3 = n C 12
⇒ n = 12 + 3 = 15
We are given that their coefficients are equal ⇒ n C 3 = n C 12 ⇒ n = 12 + 3 = 15
[ n C x = n C y ⇒ x = y (or) x + y = n]

The general term in the expansion of (a + b) n is T r+1 = nC r. a n-r. b r
∴ The general term in the expansion of (a + x) n is T r+1 = nC r. a n-r. x r
Let the three consecutive terms be r th term, (r + 1 ) th term, (r + 2 ) th term.

Given (1 + x) n
12(n – 4) = 30 + n 2 – 5n – 4n + 20 ⇒ 12n – 48 = 30 + n 2 – 9n + 20
⇒ n 2 – 9n – 12n + 50 + 48 = 0 ⇒ n 2 – 21n + 98 = 0
⇒ n 2 – 14n – 7n + 98 = 0 ⇒ n(n – 14) – 7(n – 14) = 0
⇒ (n – 7) (n – 14) = 0 ⇒ n = 7 or n = 14

This relation is true for all values of n. Equating coefficient of x n on both sides, we have
The general term in the expansion of (1 + x) 2n is T r + 1 = 2nC r (1) 2n – r. x r
Put r = n we get, T n+1 = 2nC n. x n
∴ The coefficient of x n in the expansion of (1 + x) 2n is 2nC n

(ii)
(iii) \(4\left(\frac{1}{2}\right)^{n}\)
(iv) \(\frac{(-1)^{n}}{n}\)
(v) \(\frac{2 n+3}{3 n+4}\)
(vi) 2018
The n th term a n = 2018
a 1 = 2018,
a 2 = 2018,
a 3 = 2018,
a 4 = 2018,
a 5 = 2018,
a 6 = 2018,
∴ The given sequence is 2018, 2018, 2018, 2018, 2018, 2018, ………….
This is a œnstant sequence which has same common ratio and common difference.
Hence this is an A. P, G. P and AGP.
(vii) \(\frac{3 n-2}{3^{n-1}}\)

The odd terms are a 1 = 2, a 3 = 4, a 5 = 6
The even terms are a 2 = 2, a 4 = 4, a 6 = 6
(ii)
The terms in the numerator are 1, 2, 3, 4
a = 1, d = 2 – 1 = 1
a n = a + (n – 1) d
a n = 1 + (n – 1)(1) = 1 + n – 1 = n
a n = n
The terms in the denominator are 2, 3, 4, 5, 6.
a = 2, d = 3 – 2 = 1
a n = a + (n – 1) d
a n = 2 + (n – 1) (1) = 2 + n – 1 = n + 1
a n = n + 1
∴ The n th term of the given sequence is a n = \(\frac{n}{n+1}\) for all n ∈ N
(iii)
The terms in the numerator are 1, 3, 5, 7, 9, ………….
a = 1, d = 3 – 1 = 2
a n = a + (n – 1) d
a n = 1 + (n – 1)2
a n = 1 + 2n – 2 = 2n – 1
The terms in the denominator are 2, 4, 6, 8, 10, …………..
a = 2, d = 4 – 2 = 2
a n = a + (n – 1) d
a n = 2 + (n – 1)(2)
a n = 2 + 2n – 2 = 2n
∴ The n th term of the given sequence is a n = \(\frac{2 n-1}{2 n}\) for all n ∈ N
(iv) 6, 10, 4, 12, 2, 14, 0, 16, – 2 …………………
The given sequence is 6, 10, 4, 12, 2, 14, 0, 16, – 2 ……………..
The odd terms are a 1 = 6, a 3 = 4, a 5 = 2, a 7 = 0, a 9 = – 2
∴ a n = n – 7, n is odd
The even terms are a 2 = 10, a 4 = 12, a 6 = 14, a 8 = 16
∴ a n = 8 + n, n is even.

Let the increasing numbers in G.P be \(\), a, ar.
Given \(\frac{a}{r}\) × a × ar = 5832 ⇒ a 3 = 5832 = 18 3 ⇒ a = 18
Also given \(\frac{a}{r}\), a + 6, ar + 9 form an A.P.
∴ 2(a + b) = \(\frac{a}{r}\) + (ar + 9)
⇒ (a + 6) + (a + 6) = \(\frac{a}{r}\) + (ar + 9)
⇒ (a + 6) – \(\frac{a}{r}\) = (ar + 9) – (a + 6)
⇒ a + 6 – \(\frac{a}{r}\) = ar + 9 – a – 6
⇒ a + 6 – \(\frac{a}{r}\) = ar – a + 3
Substituting the value of a = 18, we get
39r = 18r 2 + 18
18r 2 – 39r + 18 = 0
(2r – 3)(3r -2) = 0
2r – 3 = 0 or 3r – 2 = 0
r = \(\frac{3}{2}\) or r = \(\frac{2}{3}\)
Case (i) When a = 18, r = \(\frac{3}{2}\) the numbers in G.P are
Case (ii) When a = 18, r = \(\frac{2}{3}\)

The terms in the numerator are 3,5, 7
which forms an A. P with first term a = 3 and common difference d = 5 – 3 = 2
n th term t n = a + (n – 1) d
= 3 + (n – 1)(2)
= 3 + 2n – 2 = 2n + 1
t n = 2n + 1
The terms in the denominator are 1 2. 2 2, 2 2. 3 2, 3 2. 4 2 ……………….
n th term t n = n 2. (n + 1) 2

The roots are equal ⇒ ∆ = 0
(i.e.) b 2 – 4ac = 0
Hence, a = q – r; b = r – p; c = p – q
b 2 – 4ac = 0
⇒ (r – p) 2 – 4(q – r)(p – q) = 0
r 2 + p 2 – 2pr – 4[qr – q 2 – pr + pq] = 0
r 2 + p 2 – 2pr – 4qr + 4q 2 + 4pr – 4pq = 0
(i.e.) p 2 + 4q 2 + r 2 – 4pq – 4qr + 2pr = 0
(i.e.) (p – 2q + r) 2 = 0
⇒ p – 2q + r = 0
⇒ p + r = 2q
⇒ p, q, r are in A.P.

Let A be first term and R be the jmnon ratio of the G.P.
Given a = p th term of the G.P
General term of a G. P with first term A and common ratio R is t n = AR n – 1
∴ a = t p = AR P – 1
log a = log AR p-1 = log A + log R p-1 = log A + (p – 1) log R
b = q th term of the G.P
b = t q = AR q-1
log b = log AR q-1 = log A + log R q-1 = log A + (q – r)log R
c = r th term of the G.P
c = t r = AR r-1
log c = log AR r-1 = log A + log R r-1 = log A + (r – 1) log R
(q – r) log a + (r – p) log b + (p – q) log c
= (q – r) [log A + (p – 1) log R] + (r – p) [ log A + (q – 1) log R ] + (P – q) [ log A + (r – 1) log R]
= (q – r) log A + (q – r) (p – 1 ) log R + (r – p) log A + (r – p) (q – 1) log R + (P – q) log A + (p – q) (r – 1 ) log R
= [ q – r + r – p + p – q ] log A + [ (q – r) (p – 1) + (r – p) (q – 1) + (p – q)(r – 1)] log R
= 0 × log A + [pq – q – rp + r + rq – r – pq + p + pr – p – rq + q] log R
= 0 × log R = 0
∴ (q – r) log a + (r – p) log b + (p – q) log c = 0

Sum of the first n terms of an Arithmetic progression is S n = \(\) [2a + (n – 1)d]
Given S 10 = 52
52 = \(\frac{10}{2}\) [ 2a + (10 – 1) d ]
52 = 5 [2a + 9d]
52 = 10a + 45d ……….. (1)
Also given S 15 = 77
77 = \(\frac{15}{2}\) [2a + (15 – 1)d]
77 = \(\frac{15}{2}\) [2a + 14d]
77 = 15 [a + 7d]
77 = 15a + 105d ………….. (2)
Substituting the value of d in (1)

The given series is 1 + (1 + 4) + (1 + 4 + 4 2 ) + (1 + 4 + 4 2 + 4 3 ) + …………..
n th term of the series is T n = 1 + 4 + 4 2 + 4 3 + ………………. + 4 n-1
Sum to n terms of the series
S n = T 1 + T 2 + ………….. + T n

The given sequence as 1, \(\frac{4}{3}\), \(\frac{7}{9}\), \(\frac{10}{27}\), ………….
Consider the terms in the numerator 1, 4, 7, 10, ……………….. which is an Arithmetic progression
with first term a = 1, common difference d = 4 – 1 = 3
a n = a + (n – 1)d = 1 + (n – 1)3 = 1 + 3n – 3 = 3n – 2
The given sequence can be written as 1, (1 + 3)\(\left(\frac{1}{3}\right)\), (1 + 2 × 3)\(\left(\frac{1}{3}\right)^{2}\) (1 + 3 × 3) \(\left(\frac{1}{3}\right)^{3}\), …………….
where \(1, \frac{1}{3}, \frac{1}{3^{2}}, \frac{1}{3^{3}}, \ldots \ldots\) is a G. P with first term a = 1, common ratio r = \(\frac{1}{3}\)
∴ The given sequence is an Arithmetico – Geometric sequence.
which is an arithmetic – geometric sequence.
∴ The sum of first n terms of the arithmetico – geometric sequence is given by

Let the A.P. be a, a + d, a + 2d, ……..
t m + n = a + (m + n – 1)d
t m – n = a + (m – n – 1)d
t m = a + (m – 1)d
2t m = 2[a + (m – 1)d]
To prove t m + n + t m – n = 2t m
LHS t m + n + t m – n = a + (m + n – 1)d + a + (m – n – 1)d
= 2a + d [2m – 2]
= 2[a + (m – 1)d) = 2 t m = RHS

Amount of loan = 3250
Let n be the number of months taken to clear the loan
Amount paid in the first month a = 20
Increased payment in every month d = 15
∴ Amount paid in the second month = 20 + 15 = 35
Amount paid in the third month = 35 + 15 = 50
∴ The sequence of amount paid in every month is 20, 35, 50, …………. which is an A.P with first term a = 20 and common difference = 15
Given S n = \(\frac{n}{2}\) [2a + (n – 1) d]
S n = 3250
3250 = \(\frac{n}{2}\) [2 × 20 + (n – 1) 15]
6500 = n[40 + 15n – 15]
6500 = n[25 + 15n]
6500 = 25n + 15n 2
1300 = 5n + 3n 2
3n 2 + 5n – 1300 = 0
3n 2 + 65n – 60n – 1300 = 0
n(3n + 65) – 20 (3n + 65) = 0
(n – 20) (3n + 65) = 0
n = 20 = 0 or 3n + 65 = 0
n = 20 or n = \(-\frac{65}{3}\)
n = \(-\frac{65}{3}\) is not possible ∴ n = 20
Thus, in 20 months the loan is cleared.

Number of balls placed in a line = 20
Let A be the starting point.
The distance of the first ball from A = 24 m
The distance of the second ball from the first ball = 4 m
A contestant starts from point A, travels a distance of 24 m and picks the first ball and brings it back to starting point A. This is continued for each ball.
Distance travelled by the contestant to bring the first ball = 24 + 24 = 2 × 24 = 48 m
Distance travelled to bring the second ball = 2 ( 24 + 4 ) = 2 × 28 = 56 m
Distance travelled to bring the third ball = 2 × (24 + 4 + 4) = 64 m
∴ The sequence of distances travelled are 48, 56, 64 ……………
This is an Arithmetic progression with first term a = 48,
common difference d = 56 – 48 = 8,
number of terms n = 20.
The sum of 20 terms of this A.P gives the total distance travelled by the contestant in bringing all balls to the starting place
S n = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d]
S 20 = \(\frac{\mathrm{20}}{2}\) [2 × 48 + (20 – 1)8] = 10 [96 + 19 × 8]
= 10 [ 96 + 152] = 10 × 248 = 2480
Total distance traveled = 2480 m

Number of bacteria at the beginning = 30
Number of bacteria after 1 hour = 30 × 2 = 60
Number of bacteria after 2 hours = 30 × 2 2 = 120
Number of bacteria after 4 hours = 30 × 24 = 30 × 16 = 480
∴ Number of bacteria after n th hour = 30 × 2 n

The number of viruses present at the beginning = 5, Given virus, gets doubled each day
∴ The sequence of a number of viruses in each day is 5, 10, 20, 40, 80, ………………. which is a G. P with First-term a = 5, Common ratio r = \(\frac{10}{5}\) = 2
n th term t n = ar n-1
∴ On the 15 th day, the infectious virus grows over 1,50,000 units.

(ii) \(\frac{2}{(3+4 x)^{2}}\)
(iii) (5 + x 2 ) 2/3
(iv) \((x+2)^{-\frac{2}{3}}\)
Expanding binomials calculator is a free online tool that displays the expansion of the given binomial term.

(4) n(n + 1)
Explaination:
2 + 4 + 6 + ……… + 2n = 2(1 + 2 + 3 + ………….. + n)
= 2 × \(\frac{n(n+1)}{2}\)
= n(n + 1)

(1) 2
Explaination:
Given a, 8, b are in A. P ∴ 2 × 8 = a + b ⇒ a + b = 16 ——— (1)
Also a, 4, b are in G.P ∴ 4 2 = a. b ⇒ ab = 16 ——- (2)
Also a, x, b are in H.P. ∴ \(\frac{1}{a}, \frac{1}{x}, \frac{1}{b}\) are in A.P

(4) 4
Explaination:
Let a, b be the two numbers. Given A. M = \(\frac{a+b}{2}\) = 16
G.M = \(\sqrt{a b}\) = 8
a+b = 16 × 2 = 32
ab = 8 2 = 64
H.M = \(\frac{2 a b}{a+b}\)
= \(\frac{2 \times 64}{32}\) = 2 × 2 = 4

(1) d
Explaination:
= a + (n – 1) d – (a + (n – 2)d)
= a + (n – 1) d – a – (n – 2)d
= a + nd – d – a – nd + 2d = d

(1) 12
Explaination:
38 15 = (39 – 1) 15 = 39 15 – 15C 1 39 14 (1) + 15C 2 (39) 13 (1) 2 – 15C 3 (39) 12 (1) 3 ….. + 15C 14 (39) 1 (1) – 15C 15 (1)
Except -1 all other terms are divisible by 13.
∴ When 1 is added to it the number is divisible by 13. So the remainder is 13 – 1 = 12.