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find the sum of terms in nth group (1^3),(2^3,3^3),(4^3,5^3,6^3),........




d) none of these

7 years ago


Answers : (2)


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

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

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)

7 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!!!

7 years ago

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