# Match the questio

Let us assume the required sum = S

Therefore, S = 12 + 22 + 32 + … + n2

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

n3 - (n – 1)3 = 3n2 – 3n + 1

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

13 – (1 – 1)3 = 13 – 03 = 3(1)2 – 3(1) + 1

23 – (2 – 1)3 = 23 – 13 = 3(2)2 – 3(2) + 1

33 – (3 – 1)3 = 33 – 23 = 3(3)2 – 3(3) + 1

and so on

n3 - (n – 1)3 = 3n2 – 3n + 1

n3 - 03 = 3(12 + 22 + 32 + … + n2) - 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 = 13 + 23 + 33 + … + n3

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

n4 - (n – 1)4 = 4n3 – 6n2 + 4n – 1

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

14 – (1 – 1)4 = 14 – 04 = 4(1)3 – 6(1)2 + 4(1) + 1

24 – (2 – 1)4 = 24 – 14 = 4(2)3 – 6(2)2 + 4(2) + 1

34 – (3 – 1)4 = 34 – 24 = 4(3)3 – 6(3)2 + 4(3) + 1

and so on

n4 - (n – 1)4 = 4n3 – 6n2 + 4n – 1

n4 – 04 = 4(13 + 23 + 33 + … + n3) – 6(12 + 22 + 32 + … + n2) +4(1 + 2 + 3 + 4 + … + n) + (1 + 1 + 1 + 1 + … n times)  n4 = 4S – n(n + 1)(2n + 1) + 2n (n + 1) – n

4S = n4 + n + n(n + 1)(2n + 1) – 2n(n + 1)

4S = n(n3 + 1) + n(n + 1)[(2n + 1) – 2]

4S = n[n3 + 1 + (n + 1)(2n – 1)]

4S = n[n3 + 1 + 2n2 – n + 2n – 1]

4S = n[n3 + 2n2 + n]

4S = n2(n2 + 2n + 1)

4S = n2(n + 1)2 Thus, the sum of the cubes of first n natural numbers = (b) (i)

(c) Let Sn = 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 :

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