Q. 365.0( 2 Votes )

Match the questio

Answer :

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


Adding we get,


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


Adding we get,


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)



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