 # find the sum of terms in nth group (1^3),(2^3,3^3),(4^3,5^3,6^3),........ a)1\8n^3(n^2+1)(n^2+3) b)1\16n^3(n^2+16)(n^2+12) c)1\12n^3(n^2+2)(n^2+4) d) none of these

### find the sum of terms in nth group (1^3),(2^3,3^3),(4^3,5^3,6^3),........a)1\8n^3(n^2+1)(n^2+3)b)1\16n^3(n^2+16)(n^2+12)c)1\12n^3(n^2+2)(n^2+4)d) none of these AskIITians Expert Hari Shankar IITD
17 Points
13 years ago

Lets first find out the first and last terms of the nth group.

The first terms are cubes of 1,2,4,7,....tk

Sum of k terms = 1+2+4+7+....+tk......................(1)

Sum of (k-1) terms = 1+2+4+7+....+tk-1...............(2)

Subtracting (2) from (1),

Sum of k terms - Sum of (k-1) terms = 1+(2-1)+(4-2)+(7-4)+.....+(tk-tk-1) = 1+1+2+3+...upto k terms = 1+ (Sum of first (k-1) natural numbers

= 1 + (k(k-1))/2

But we also know that Sum of k terms - Sum of (k-1) terms is the kth term. So, kth term is 1 + (k(k-1))/2.

Now consider the last terms.

The last terms are First term + k.

=1 + (k(k-1))/2 + k

So, the nth set is [(1 + (n(n-1))/2)3, (1 + (n(n-1))/2 + 1)3,.....(1 + (n(n-1))/2+n)3]

For example, the fourth set is [(1 + (4(4-1))/2)3, (1 + (4(4-1))/2 + 1)3,.....(1 + (4(4-1))/2+4)3]

= [73,83,93,103]

Sum of terms of fourth set = Sum of cubes upto 103 - Sum of cubes upto 63

Similarly, Sum of terms of nth set = Sum of cubes upto (1 + (n(n-1))/2 + n)3 - Sum of cubes upto (1 + (n(n-1))/2 - 1 )3

= Sum of cubes upto ((n2+n+2)/2)3 - Sum of cubes upto ((n2-n)/2 )3

Now apply the sum of cubes formulas to both these terms and simplify.
Sum of cubes upto ((n2+n+2)/2)3 = [ {(n2+n+2)/2} {(n2+n+2)/2 + 1}/2 ]

Sum of cubes upto ((n2-n)/2 )3 = [ {(n2-n)/2} {(n2-n)/2 + 1}/2 ]2

Then the final answer will be [ {(n2+n+2)/2} {(n2+n+2)/2 + 1}/2 ]2 - [ {(n2-n)/2} {(n2-n)/2 + 1}/2 ]

(You will have to further simplify this)

13 years ago

find the sum of first n-1 natural numbers that is (n-1)n/2.

this implies the first term of the nth term would be (n-1)n/2+1.

hence find the sum of the terms in the nth term.that is {the sum of cubes of natural number till (n-1)n/2+n}-{the sum of cubes of natural number till (n-2)(n-1)/2+n-1}.

the answer turns out to be a. Use the formulae for the sum of cubes of first n natural no. to be (n(n+1)/2)2.

ask me if u dont get the answer.its a bit calculative!!!