Consider the following sequence of sums:
1 , 2 + 3 , 4 + 5 + 6 , 7 + 8 + 9 + 1 0 , . . .
Which of the following is the sum of the 23rd term?
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Very nice solution! Well done!
According to sequence 23rd term will contain 23 numbers. Since 1st term contain 1 no., 2nd contain 2 no. so it can be treated like another sequence of 1,2,3,4.....up to...22nd term. This sequence will give the numbers already used up to 22nd term. The 22nd term ends at 253, so the 23rd term will start from 254 and end to 276
This is the logical way to solve this question, as a kind of 'Spot the pattern' approach. Well Done! But can you solve this by finding the n t h term formula?
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2 1 n ( n 2 + 1 )
the sum of nth term in this series is = (n^3+n)/2
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It is noted that the last element M n of n t h term, such as 1 , 3 , 6 , 1 0 . . . , is given by M n = 2 n ( n + 1 ) .
Therefore, the sum up to the n t h term is given by: S n = 2 M n ( M n + 1 ) .
The sum of the 2 3 r d term is s 2 3 = S 2 3 − S 2 2 = 2 M 2 3 ( M 2 3 + 1 ) − M 2 2 ( M 2 2 + 1 )
M 2 3 = 2 2 3 ( 2 3 + 1 ) = 2 7 6
M 2 2 = 2 2 2 ( 2 2 + 1 ) = 2 5 3
Therefore, s 2 3 = 2 2 7 6 × 2 7 7 − 2 5 3 × 2 5 4 = 2 7 6 4 5 2 − 6 4 2 6 2 = 2 1 2 1 9 0 = 6 0 9 5