Q. 365.0( 2 Votes )

# Match the questions given under Column I with their appropriate answers given under the Column II.

Answer :

Let us assume the required sum = S

Therefore, S = 1^{2} + 2^{2} + 3^{2} + … + n^{2}

Now, we will use the below identity to find the value of S:

n^{3} - (n – 1)^{3} = 3n^{2} – 3n + 1

Substituting, n = 1, 2, 3, 4, 5,…, n in the above identity, we get

1^{3} – (1 – 1)^{3} = 1^{3} – 0^{3} = 3(1)^{2} – 3(1) + 1

2^{3} – (2 – 1)^{3} = 2^{3} – 1^{3} = 3(2)^{2} – 3(2) + 1

3^{3} – (3 – 1)^{3} = 3^{3} – 2^{3} = 3(3)^{2} – 3(3) + 1

and so on

n^{3} - (n – 1)^{3} = 3n^{2} – 3n + 1

Adding we get,

n^{3} - 0^{3} = 3(1^{2} + 2^{2} + 3^{2} + … + n^{2}) - 3(1 + 2 + 3 + 4 + … + n) + (1 + 1 + 1 + 1 + … n times)

Taking common n(n + 1), we get

Thus, the sum of the squares of first n natural numbers =

∴(a) ↔ (iii)

(b) Let us assume the required sum = S

Therefore, S = 1^{3} + 2^{3} + 3^{3} + … + n^{3}

Now, we will use the below identity to find the value of S:

n^{4} - (n – 1)^{4} = 4n^{3} – 6n^{2} + 4n – 1

Substituting, n = 1, 2, 3, 4, 5,…, n in the above identity, we get

1^{4} – (1 – 1)^{4} = 1^{4} – 0^{4} = 4(1)^{3} – 6(1)^{2} + 4(1) + 1

2^{4} – (2 – 1)^{4} = 2^{4} – 1^{4} = 4(2)^{3} – 6(2)^{2} + 4(2) + 1

3^{4} – (3 – 1)^{4} = 3^{4} – 2^{4} = 4(3)^{3} – 6(3)^{2} + 4(3) + 1

and so on

n^{4} - (n – 1)^{4} = 4n^{3} – 6n^{2} + 4n – 1

Adding we get,

n^{4} – 0^{4} = 4(1^{3} + 2^{3} + 3^{3} + … + n^{3}) – 6(1^{2} + 2^{2} + 3^{2} + … + n^{2}) +4(1 + 2 + 3 + 4 + … + n) + (1 + 1 + 1 + 1 + … n times)

⇒ n^{4} = 4S – n(n + 1)(2n + 1) + 2n (n + 1) – n

⇒ 4S = n^{4} + n + n(n + 1)(2n + 1) – 2n(n + 1)

⇒ 4S = n(n^{3} + 1) + n(n + 1)[(2n + 1) – 2]

⇒ 4S = n[n^{3} + 1 + (n + 1)(2n – 1)]

⇒ 4S = n[n^{3} + 1 + 2n^{2} – n + 2n – 1]

⇒ 4S = n[n^{3} + 2n^{2} + n]

⇒ 4S = n^{2}(n^{2} + 2n + 1)

⇒ 4S = n^{2}(n + 1)^{2}

Thus, the sum of the cubes of first n natural numbers =

∴(b) ↔ (i)

(c) Let S_{n} = 2 + 4 + 6 + … + 2n

= 2(1 + 2 + 3 + … + n)

= n(n + 1)

∴ (a) ↔ (ii)

(d) Let S be the required sum.

Therefore, S = 1 + 2 + 3 + 4 + 5 + … + n

Clearly, it is an Arithmetic Progression whose first term = 1, last term = n and number of terms = n.

Using the formula,

Therefore,

or we can say that,

∴ (d) ↔ (iv)

Rate this question :

A person has 2 parents, 4 grandparents, 8 great grand parents, and so on. Find the number of his ancestors during the ten generations preceding his own.

RD Sharma - MathematicsThe sum of the series to n terms is

RD Sharma - MathematicsHow many terms of the G.P. are needed to give the sum ?

RD Sharma - MathematicsIf , then S_{n} is equal to

Let a_{n} be the nth term of the G.P. of positive numbers. Let and such that α ≠ β. Prove that the common ratio of the G.P. is α /β

Find the sum of 2n terms of the series whose every even term is ‘a’ times the term before it and every odd term is ‘c’ times the term before it, the first term being unity.

RD Sharma - MathematicsIf , find the value of .

RS Aggarwal - MathematicsFind the sum (41 + 42 + 43 + …. + 100).

RS Aggarwal - MathematicsFind the sum (2 + 4 + 6 + 8 +… + 100).

RS Aggarwal - MathematicsIf, prove that,

.

RS Aggarwal - Mathematics