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Flag Magical Mathematics[Interesting Approach]> If 1^3 =1, 2^3=3+5, 3^3=7+9+11, 4^3=13+15...
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If 1^3 =1, 2^3=3+5, 3^3=7+9+11, 4^3=13+15+17+19........, then write the nth term of the series (as a similar series )

Aishwarya , 9 Years ago
Grade 12
anser 1 Answers
arun

Last Activity: 9 Years ago

considering them as a set i.e. 1^3 is a set containing term, 2^3 is a set containing 2 terms, similarly n^3 is a set containing n terms 
then total numbers of terms upto n^3 set is \frac{n(n+1)}{2}
similarly, total number of terms upto (n-1)^3 is \frac{n(n-1)}{2}
as the series in this is an A.P. with common difference of 2
 so, nth term for this is \left ( 1 \right ) + \left ( n-1 \right )\left ( 2 \right ) = 2n-1
this means  \frac{n(n+1)}{2}th term in it is 2(\frac{n(n+1)}{2}) -1 = n^2 +n -1
this means the last term in the set n^3 is n^2 +n -1
now first term in set n^3 will be \frac{n(n-1)}{2} +1th term 
which is 2\left ( \frac{n(n-1)}{2}+1 \right )-1=n^2 -n+1
so the nth term of the series is 
n^3 = (n^2 -n+1) + (n^2 -n +1 +2) + ..... + (n^2 +n-1)

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