Some special
1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 3 1 0 3 ⋮
Find the sum of the numbers in the 5 0 th line of the above progression.
The answer is 97812546875.
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1 solution
How do we calculate the first part?
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Starting from 1, first number of the second row is 1 + ( 1 ) . Similarly, first number of third row is 1 + ( 1 + 2 ) and fourth row is 1 + ( 1 + 2 + 3 ) and so on. First number of 5 0 th row is 1 + ( 1 + 2 + 3 + . . . + 4 9 ) .
Just to clarify you should state the formula which is (n(n+1)/2)^2 or ((n^2)+n/2)^2.
Each set contains 1 more than the number of terms of the previous set. Thus, the sum of the 50 sets will have an arithmetic sum of 1 2 7 5 terms. We want to find the sum of the last 50 terms. Since all the numbers are consecutive cubes, we can find the sum of the 50th set by simply subtracting:
( 1 3 + 2 3 + 3 3 + . . . + 1 2 7 5 3 ) − ( 1 3 + 2 3 + 3 3 + . . . + 1 2 2 5 3 )
And by using the sum of first n cubes formula we get the answer.