NCERT Exemplar: Sequence and Series | Mathematics (Maths) for JEE Main & Advanced PDF Download

Q.1. The first term of an A.P.is a, and the sum of the first p terms is zero, show that the sum of its next q terms is
[Hint: Required sum = S_{p + q} – S_p]
Ans.
Given that a₁ = a and S_p = 0
Sum of next q terms of the given A.P. = S_p+q – S_p
∴
and
⇒ 2a + (p – 1)d = 0 ⇒ (p – 1)d = – 2a
⇒
Sum of next q terms = S_p+q – S_p


Hence, the required sum =

Q.2. A man saved Rs 66000 in 20 years. In each succeeding year after
the first year he saved Rs 200 more than what he saved in the previous
year. How much did he save in the first year?
Ans.
Let ₹ x  be saved in first year.
Annual increment = ₹ 200
which forms an A.P.
first term = a and common difference d = 200
n = 20 years
∴
⇒ 66000 = 10[2a + 3800] ⇒ 6600 = 2a + 3800
⇒ 2a = 6600 - 3800 ⇒ 2a = 2800 ⇒ a = 1400 Hence, the man saved ₹1400 in the first year.

Q.3. A man accepts a position with an initial salary of Rs 5200 per month. It is understood that he will receive an automatic increase of Rs 320 in the very next month and each month thereafter.
(a)  Find his salary for the tenth month
(b)  What is his total earnings during the first year?
Ans.
Given that fixed increment in the salary of a man
=₹ 320 each
month Initial salary = ₹ 5200 which makes an A.P.
whose first term (a) = ₹ 5200 and common difference (d) = ₹ 320
(i) Salary for the tenth month
a₁₀ = a + (n – 1)d
= 5200 + (10 – 1) x 320 = 5200 + 2880 = ₹ 8080
(ii) Total earning during the first year (12 months)

= 6[10400 + 3520] = 6 x 13920 = ₹ 83520
Hence, the required amount is (i) ₹ 8080 (ii) ₹ 83520.

Q.4. If the pth and qth terms of a G.P. are q and p respectively, show that its (p + q)^th term is
Ans.
Let a be the first term and r be the common ratio of a G.P.
Given that a_p = q ⇒ ar^p–1 = q ...(i)
and a_q = p ⇒ ar^q–1 = p ...(ii)
Dividing eq. (i) by eq. (ii) we get,

putting the value of r in eq. (i), we get


Now T_p+q = ar^p+q–1 =


Hence, the required term =

Q.5. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?
Ans.
Here, first term a = 5 and the common difference d = 2
let the carpenter will take n days to finish the job
S_n = 192

⇒ 192 x 2 = n[10 + 2n - 2] ⇒ 384 = n(2n + 8)
⇒ 384 = 2n² + 8n ⇒ 2n² + 8n - 384 = 0
⇒ n² + 4n – 192 = 0 ⇒ n² + 16n – 12n – 192 = 0
⇒ n(n + 16) – 12 (n + 16) = 0 ⇒ (n - 12) (n + 16) = 0
⇒ n = 12    [∵ n ≠ –16]
Hence, the required number of days = 12.

Q.6. We know the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
Ans.
Since, the sum of all interior angles of a polygon of n sides
= (2n – 4) x 90°
∴ Sum of interior angles of a polygon of sides 3
= (2 x 3 - 4) x 90° = 180°
Sum of interior angles of a polygon of sides 4
= (2 x 4 - 4) x 90° = 360°
Similarly, the sum of interior angles of the polygon of sides,     5, 6, 7 … are 540°, 720°, 900°…
Therefore, the series will be 180°, 360°, 540°, 720°, 900°…
which is A.P.
Here a = 180°, d = 180°
We have to find the sum of interior angles of a polygon of  21 sides i.e. 19th term
a_n = a + (n – 1)d
a₁₉ = 180° + (19 – 1)180° = 180° + 18 x 180°
= 180° + 3240° = 3420°
Hence, the required sum of interior angles = 3420°.

Q.7. A side of an equilateral triangle is 20cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.
Ans.
The side of the first equilateral DABC = 20 cm
By joining the mid points of the sides of this triangle, we get the second  equilateral     triangle which each side
  
[∴ The line joining the mid-points of two sides of a triangle
is 1/2 and parallel to third side of the triangle] 
Similarly each side of the third equilateral triangle =  
∴ Perimeter of first triangle = 20 x 3 = 60 cm
Perimeter of the second triangle = 10 x 3 = 30 cm
and the perimeter of the third triangle = 5 x 3 = 15 cm
Therefore, the series will be 60, 30, 15,
which is G.P. in which a = 60, and r =   
Now, we have to find the perimeter of the sixth inscribed equilateral triangle     
∴ a₆ = ar^6–1
=
Hence, the required perimeter = 

Q.8. In a potato race 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?  
Ans.
As per the given information we have the following diagram

Starting point = S Distance traveled to bring the first potato = 24 + 24 = 48 m Distance traveled to bring the second potato = 2(24 + 4) = 56 m
Distance traveled to bring the third potato = 2(24 + 4 + 4) = 64 m     Therefore, the series will be = 48, 56, 64, …
which an A.P. in which a = 48, d = 56 – 48 = 8
We have to find  the  total distance to bring all the potatoes
back, so, n = 20
∴
⇒
= 10 x 248 = 2480 m
Hence, the required distance = 2480 m

Q.9. In a cricket tournament 16 school teams participated. A sum of Rs 8000 is to be awarded among themselves as prize money. If the last placed team is awarded Rs 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?
Ans.
Let the prize amount got by first place team be ₹ a
Since, the prize money increases by the same amount for     successive finishing places, therefore the series will be A.P.
∴ a_n = 275, n = 16 and S₁₆ = 8000
a_n = a + (n – 1)d
275 = a + (16 – 1)(– d)
[∴ Common difference d is (-) as the series is decreasing]
⇒ 275 = a - 15d     ...(i)

⇒ 8000 = 8[2a – 15d] ⇒ 2a - 15d = 1000 ...(ii)
Solving eq. (i) and eq. (ii) we get
a = 725 and d = 30
Hence, the required award received by first place term =₹ 725.  

Q.10. If a₁, a₂, a₃, ..., a_n are in A.P., where a_i > 0 for all i, show that

Ans.
Given that a₁, a₂, a₃, a₄, …, a_n are in A.P.
∴ Common difference d = a₂ - a₁ = a₃ - a₂ = a₄ - a₃ = ... = a_n - a_n-1


Adding the above terms, we get


Now     a_n = a₁ + (n – 1)d
⇒ a_n – a₁ = (n – 1)d
...(ii)
From Eq. (i) and eq. (ii) we get

Hence proved.

Q.11. Find the sum of the series
(3³ – 2³) + (5³ – 4³) + (7³ – 6³) + ... to (i) n terms (ii) 10 terms
Ans.
Given series
⇒ (3³ - 2³) + (5³ - 4³) + (7³ - 6³) + ...
= (3³ + 5³ + 7³ + …) – (2³ + 4³ + 6³ + …)
⇒ [3³ + 5³ + 7³ + ... (2n + 1)³] - [2³ + 4³ + 6³ + ... (2n)³]
∴ Tn = (2n + 1)³ – (2n)³ 
= (2n + 1 - 2n) [(2n + 1)² + (2n + 1) (2n) + (2n)² ]
[∴ a³ - b³ = (a - b) (a² + ab + b²)]
= 1 × [4n² + 1 + 4n + 4n² + 2n + 4n²]
= 12n² + 6n + 1
(i)

= 2n(n + 1) (2n + 1) + 3n(n + 1) + n
= n[2(n + 1) (2n + 1) + 3(n + 1) + 1]
= n[2(2n² + 3n + 1) + 3n + 3 + 1]
= n[4n² + 6n + 2 + 3n + 4]
= n[4n² + 9n + 6]
= 4n³ + 9n² + 6n
(ii) S₁₀ = 4(10)³ + 9(10)² + 6(10) = 4 x 1000 + 900 + 60
= 4000 + 960 = 4960

Q.12. Find the r^th term of an A.P. sum of whose first n terms is 2n + 3n².
[Hint: a_n = S_n – S_n–1]
Ans.
Given that S_n = 2n + 3n²
⇒ S₁ = 2 x 1 + 3(1)² = 5
⇒ S₂ = 2 x 2 + 3 x 4 = 16
⇒ S₃ = 2 x 3 + 3 x 9 = 33

…     … …

∴ S₁ = a₁ = 5
S₂ - S₁ = a₂ = 16 - 5 = 11
∴ d = a₂ – a₁ = 11 – 5 = 6
Now T_r = a₁ + (r – 1)d
= 5 + (r – 1)6 = 5 + 6r – 6 = 6r – 1
Hence, the required r^th term is 6r – 1

Long Answer Type
Q.13.  If A is the arithmetic mean and G₁, G₂ be two geometric means between any two numbers, then prove that

Ans.
Let the two numbers be x and y
∴ ...(i)
If  G₁ and G₂ be the geometric means between x and y then x, G₁, G₂, y are in G.P.
Then  y = xr^4–1 [∴ a_n = ar_n–1]

Now    
and

= x + y = 2A LHS. [using eq. (i)]
∴ LHS = RHS 
Hence proved.

Q.14. If θ₁, θ₂, θ₃, ..., θ_n are in A.P., whose common difference is d, show that secθ₁ secθ₂ + sec_θ2 secθ₃ + ... + secθ_n–1 secθ_n
Ans.
Since θ₁, θ₂, θ₃, …, θ_n are in A.P.
∴ θ₂ – θ₁ = θ₃ – θ₂ = … = θ_n – θ_n–1 = d
Now we have to prove that
sec θ₁ × sec θ₂ + sec θ₂ ×  sec θ₃ + ... + sec θ_n-1 . sec θ_n

Taking only

Similarly we can solve other terms which will be

Here LHS =

LHS = RHS
Hence proved.

Q.15. If the sum of p terms of an A.P. is q and the sum of q terms is p, show that the sum of p + q terms is – (p + q). Also, find the sum of first p – q terms (p > q).
Ans.
Let a be the first term and d the common difference of the given A.P.

Subtracting eq. (ii) from eq. (i) we get

Substituting the value of d in eq. (i) we get 
   


Hence proved.

Q.16. If p^th, q^th, and r^th terms of an A.P. and G.P. are both a, b and c respectively, show that a^b–c . b^{c – a} . c^{a – b} = 1
Ans.
Let A and d be the first term and common difference respectively of an A.P. and x and R be the first term and common ratio respectively of  the G.P.         
∴ A + (p – 1)d = a ...(i)
A + (q – 1)d = b ...(ii)
and A + (r – 1)d = c ...(iii)
For G.P., we have
xR^p–1 = a ...(iv)
xR^q–1 = b ...(v)
and xR^r–1 = c ...(vi)
Subtracting eq. (ii) from eq. (i) we get
(p – q)d = a – b ...(vii)
Similarly, (q – r)d = b – c ...(viii)
and (r – p)d = c – a ...(ix)
Now we have to prove that
a^b–c × b^c–a × c^a–b = 1
L.H.S.  a^b–c × b^c–a × c^a–b
= [ xR ^{p -1} ]^{( q - r) d} × [xR ^q-1 ]^{( r- p)d}× [ xR ^{r -1} ]^{( p - q)}d [from (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix)]
= x^(q-r)d × R^{(p-1) (q-r)d} × x^(r-p)d × R^{(q-1) (r-p)d} × x^(p-q)d × R^{(r-1) (p-q)d}
= x^{(q-r)d + (r-p)d + (p-q)d} R^{(p-1)(q-r)d + (q-1) (r-p)d + (r-1) (p-q)d}
= x^{(q-r+r-p+p-q)d} × R^{(pq-pr-q+r+qr-pq-r+p+pr-qr-p+q)d}
= x^(0)d × R^(0)d = x⁰ × R⁰ = 1 R.H.S.     
L.H.S. = R.H.S.
Hence proved.

Objective Type Questions
Q.17. If the sum of n terms of an A.P. is given by S_n = 3n + 2n², then the common difference of the A.P. is
(a) 3
(b) 2
(c) 6
(d) 4
Ans. (d)
Solution.
Given that S_n = 3n + 2n²
S₁ = 3(1) + 2(1)² = 5
S₂ = 3(2) + 2(4) = 14
S₁ = a₁ = 5
S₂ – S₁ = a₂ = 14 – 5 = 9
∴ Common difference d = a₂ - a₁ = 9 - 5 = 4
Hence, the correct option is (d).

Q.18. The third term of G.P. is 4. The product of its first 5 terms is
(a) 4³
(b) 4⁴
(c) 4⁵
(d) None of these
Ans. (c)
Solution.
Given that T₃ = 4
⇒ ar^3–1 = 4    [∴ T_n = ar^n-1]
⇒ ar² = 4
Product of first 5 terms = a × ar × ar² × ar³ × ar⁴
= a⁵r¹⁰ = (ar²)⁵ = (4)⁵
Hence, the correct option is (c).

Q.19. If 9 times the 9^th term of an A.P. is equal to 13 times the 13^th term, then the 22^nd term of the A.P. is
(a) 0
(b) 22
(c) 220
(d) 198
Ans. (a)
Solution.
T_n = a + (n – 1)d
∴ T₉ = a + 8d
and T₁₃ = a + 12d
As per the given condition
9[a + 8d] = 13[a + 12d]
⇒ 9a + 72d = 13a + 156d ⇒ - 4a = 84d
⇒ a = – 21d ...(i)
Now T₂₂ = a + 21d = – 21d + 21d = 0 [from eq. (i)]
Hence, the correct option is (a).

Q.20. If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.P. then the common ratio of  the G.P. is
(a) 3
(b)1/3
(c) 2
(d)1/2
Ans. (b)
Solution.
Since x, 2y, 3z are in A.P.
∴ 2y – x = 3z –2y
⇒ 4y = x + 3z ...(i)
Now  x, y, z are in G.P.
∴ Common ratio r =
∴ y² = xz
putting the value of x from eq. (i), we get
 y² = (4y – 3z)z ⇒ y² = 4yz – 3z²
⇒ 3z² – 4yz + y2 = 0 ⇒ 3z² - 3yz - yz + y² = 0
⇒ 3z(z - y) - y(z - y) = 0 ⇒ (3z - y) (z - y) = 0
⇒ 3z – y = 0 and z – y = 0
⇒ 3z = y and z ≠ y
[∴ z and y are distinct numbers]
⇒ (from eq. (ii))
Hence, the correct option is (b).

Q.21. If in an A.P., S_n = q n² and S_m = qm², where Sr denotes the sum of r terms of the A.P., then S_q equals
(a)q³/2
(b) mnq
(c) q³
(d) (m + n) q²
Ans. (c)
Solution.
The given series is A.P. whose first term is a and common difference is d
∴
⇒ 2a + (n – 1)d = 2qn     ...(i)

⇒ 2a + (m – 1)d = 2qm     ...(ii)
Solving eq. (i) and eq. (ii) we get

(m – n)d = 2qm – 2qn
(m – n)d = 2q(m – n)
∴ d = 2q
Putting the value of d in eq. (ii) we get    
2a + (m – 1) × 2q = 2qm ⇒ 2a = 2qm - (m -1)2q
⇒ 2a = 2q(m - m + 1) ⇒ 2a = 2q ⇒ a = q

Hence, the correct option is (c).

Q.22. Let S_n denote the sum of the first n terms of an A.P.
If S_2n = 3S_n then S_3n : S_n is equal to
(a) 4
(b) 6
(c) 8
(d) 10
Ans. (b)
Solution.

∴

As per the condition of the question, we have
S_2n = 3 × S_n
⇒
⇒ 2[2a + (2n – 1)d] = 3[2a + (n – 1)d]
⇒ 4a + (4n – 2)d = 6a + (3n – 3)d
⇒ 6a + (3n – 3)d – 4a – (4n – 2)d = 0
⇒ 2a + (3n – 3 – 4n + 2)d = 0 ⇒ 2a + (- n - 1)d = 0
⇒ 2a – (n + 1)d = 0 ⇒ 2a = (n + 1)d (i)

               
Hence, the correct option is (b).

Q.23. The minimum value of 4^x + 4^1–x , x ∈ R, is
(a) 2
(b) 4
(c) 1 
(d) 0
Ans. (b)
Solution.
We know that AM ≥ GM

Hence, the correct option is (b).

Q.24. Let S_n denote the sum of the cubes of the first n natural numbers and s_n denote the sum of the first n natural numbers. Thenequals
(a)
(b)
(c)
(d)None of these
Ans. (a)
Solution.
Given that
Let T_n be the nth term of the above series

Now sum of the given series

Hence, the correct option is (a).

Q.25. If t_n denotes the nth term of the series 2 + 3 + 6 + 11 + 18 + ... then t⁵⁰ is
(a) 49² – 1
(b) 49²
(c) 50² + 1
(d) 49² + 2
Ans. (d)
Solution.
Let S_n = 2 + 3 + 6 + 11 + 18 + … + t₅₀
Using method of difference, we get
S_n = 2 + 3 + 6 + 11 + 18 + … + t₅₀ ...(i)
and S_n = 0 + 2 + 3 + 6 + 11 + … + t₄₉ + t₅₀ ...(ii)
Subtracting eq. (ii) from eq. (i), we get
0 = 2 + 1 + 3 + 5 + 7 + … – t₅₀ terms
⇒ t₅₀ = 2 + (1 + 3 + 5 + 7 +  …  upto 49 terms)

Hence, the correct option is (d).

Q.26. The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is 216 cm³ and the total surface area is 252 cm². The length of the longest edge is
(a) 12 cm
(b) 6 cm
(c) 18 cm
(d) 3 cm
Ans. (a)
Solution.
Let the length, breadth and height of a rectangular block be
 a and ar. [Since they are is G.P] 
∴ Volume = l x b x h 

⇒ a³ = 216 ⇒ a = 6
Now total surface area  = 2[lb + bh + lh]


⇒ 2r² – 5r + 2 = 0 ⇒ 2r² – 4r – r + 2 = 0
⇒ 2r(r – 2) –1(r – 2) = 0 ⇒ (r - 2) (2r - 1) = 0
⇒ r – 2 = 0 and 2r – 1 = 0
∴
Therefore, the three edge are:
If r = 2 then edges are 3, 6, 12
If r =then edges are 12, 6, 3
So, the length of the longest edge = 12
Hence, the correct option is (a).

Fill in the blanks 
Q.27. For a, b, c to be in G.P. the value ofis equal to .............. .
Ans.
Since a, b and c are in G.P
∴
⇒ b = ar and c = br ⇒ c = ar × r = ar²

Hence, the correct value of the filler is

Q.28. The sum of terms equidistant from the beginning and end in an A.P. is equal to ............ .
Ans.
Let A.P be a, a + d, a + 2d, a + 3d, …, a + (n – 1)d
Taking first and last term
a₁ + a_n = a + a + (n – 1)d = 2a + (n – 1)d
Taking second and second last term     
a₂ + a_n–1 = (a + d) + [a + (n – 2)d] = 2a + (n – 1)d = a₁ + a_n
Taking third from the beginning and the third from the end
a₃ + a_n–2 = (a + 2d) + [a + (n – 3)d] = 2a + (n – 1)d = a₁ + a_n
From the above pattern, we observe that the sum of terms     
equidistant from the beginning and the end in an A.P is equal to the
[first term + last term]
Hence, the correct value of the filler is first term + last term.    

Q.29. The third term of a G.P. is 4, the product of the first five terms is ................ .
Ans.
Given     T₃ = 4
∴ ar² = 4    ...(i)
Product of first five terms = a × ar × ar² × a^r3 × ar⁴
= a⁵r¹⁰ = (ar²)⁵ = (4)⁵ [from eq. (i)]
Hence, the correct value of the fillter is (4)5.

TRUE/FALSE
Q.30. Two sequences cannot be in both A.P. and G.P. together.
Ans.
Let us consider aG.P, a , ar and ar²
If it is in A.P then ar – a ≠ ar² – ar
Hence, the given statement is True.

Q.31. Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.
Ans.
Let us consider a sequence of prime numbers 2, 3, 5, 7, 11, … It is clear that this progression is a sequence but sequence is not a progression because it does not follow a specific pattern.
Here, the given statement is True.

Q.32. Any term of an A.P. (except first) is equal to half the sum of terms which are equidistant from it. 
Ans.
Let us consider an A.P a, a + d, a + 2d, …
∴ a₂ + a₄ = a + d + a + 3d = 2a + 4d = 2a₃
⇒

⇒ = a + 3d = a₄      
Hence, the given statement is True.

Q.33. The sum or difference of two G.P.s, is again a G.P.
Ans. Let us consider two G.P.’s
a₁ , a₁r₁ , a₁r₁² , a₁r₁³ .... a₁r₁^n-1
and  a₂ , a₂ r₂ , a₂ r₂² , a₂ r₂³ ,.... a₂ r₂^n-1
Now Sum of  two G.Ps....    
(a₁ + a₂ ) + (a₁r₁ + a₂ r₂ ) + (a₁r₁² + a₂ r₂² ).....

Now let us consider the difference G.P’s
( a₁ - a₂ ) + ( a₁r₁ - a₂ r₂ ) + ( a₁r₁² - a₂ r₂² )

Hence, the given statement is False.

Q.34. If the sum of n terms of a sequence is quadratic expression then it always represents an A.P.
Ans.
Let S_n = an² + bn + c (quadratic expression)
S₁ = a + b + c
∴ a₁ = a + b + c
S₂ = 4a + 2b + c
a₂ = S₂ – S₁ = (4a + 2b + c) – (a + b + c) = 3a + b
S₃ = 9a + 3b + c
⇒ a₃ = S₃ – S₂ = (9a + 3b + c) – (4a + 2b + c) = 5a + b
Common difference d = a₂ – a₁ = (3a + b) – (a + b + c)
= 2a – c
and d = a₃ – a₂ = (5a + b) – (3a + b) = 2a
Here, we observe that a₂ – a₁ ≠ a₃ – a₂
So it does not represent an A.P
Hence, the given statement is False.

Q.35. Match the questions given under Column I with their appropriate answers given under the Column II.
 
Ans.

Hence it is G.P
∴ (a) ↔ (iii)
(b) 2, 3, 5, 7
Here a₂ – a₁ = 3 – 2 = 1
a₃ – a₂ = 5 – 3 = 2
∴ a₂ – a₁ ≠ a₃ – a₂
Hence it is not A.P
(c) 13, 8, 3, – 2, – 7
Here a₂ – a₁ = 8 – 13 = – 5
a₃ – a₂ = 3 – 8 = – 5
So, a₂ – a₁ = a₃ – a₂ = – 5
So, it is an A.P
Hence (c) ↔ (i)

Q.36. 

Ans.
(a) Let S_n = 12 + 22 + 32 + … + n²
K₃ – (K – 1)³ = 3K² – 3K + 1
For K = 1, 1³ – 0³ = 3(1)² – 3(1) + 1
For K = 2, 2³ – 1³ = 3(2)² – 3(2) + 1
For K = 3, 3³ – 2³ = 3(3)² – 3(3) + 1
For K = n, n³ – (n – 1)³ = 3(n)² – 3(n) + 1
Adding Column wise, we get
n³ – 0³ = 3(1² + 2² + 3² + … + n²)
– 3(1 + 2 + 3 + … +n) + n

⇒ 6 × S_n = 2n³ + 3n² + n
⇒ 6 × S_n = n(2n² + 3n + 1)
⇒ 6 × S_n = n[2n² + 2n + n + 1]
⇒ 6 × S_n = n(n + 1) (2n + 1)
⇒
Here (a) ↔ (iii)
(b) Let S_n = 1³ + 2³ + 3³ + … + n³ 
K⁴ – (K – 1)⁴ = 4K³ – 6K² + 4K – 1
For K = 1, 1⁴ – 0⁴ = 4(1)³ – 6(1)² + 4(1) – 1
For K = 2, 2⁴ – 1⁴ = 4(2)³ –6(2)² + 4(2) – 1
For K = 3, 3⁴ – 2⁴ = 4(3)3 – 6(3)² + 4(3) – 1
For K = n, n⁴ – (n –1)⁴ = 4(n)³ – 6(n²) + 4(n) – 1
Adding column wise, we get  
⇒ n⁴ – 0⁴ = 4(1³ + 2³ + 3³ + … n³)
– 6(1² + 2² + 3² + … + n²)
+ 4(1 + 2 + 3 + … n) – n

⇒ n⁴ = 4 × S_n - n(n + 1) (2n + 1) + 2n(n + 1) - n
⇒ n⁴ + n(n + 1) (2n + 1) - 2n(n + 1) + n = 4 × S_n
⇒ n    [n³ + (n + 1) (2n + 1) - 2n - 2 + 1] = 4 × S_n 
⇒ n [n³ + 2n² + 3n + 1 – 2n – 1] = 4 × S_n 
⇒ n[n³ + 2n² + n] = 4 × S_n
⇒
⇒
∴
Hence (b) ↔ (i)
(c) Let  S_n = 2  +  4 + 6 + ....+ 2n
= 2(1 + 2 + 3 + .... + n)

= n(n + 1)
Hence (c) ↔ (ii)
(d) Let S_n = 1 + 2 + 3 +...+ n =
Hence (d) ↔ (iv)