Stewart's Sequences 2 of 5

Consider the following sequence of sums:

1 , 2 + 3 , 4 + 5 + 6 , 7 + 8 + 9 + 10 , . . . 1,\quad 2+3,\quad 4+5+6,\quad 7+8+9+10,\quad ...

Which of the following is the sum of the 23rd term?

6075 6095 6105 6085

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3 solutions

It is noted that the last element M n M_n of n t h n^{th} term, such as 1 , 3 , 6 , 10... 1, 3, 6, 10... , is given by M n = n 2 ( n + 1 ) M_n = \dfrac {n}{2} (n+1) .

Therefore, the sum up to the n t h n^{th} term is given by: S n = M n 2 ( M n + 1 ) S_n = \dfrac {M_n}{2} (M_n+1) .

The sum of the 2 3 r d 23^{rd} term is s 23 = S 23 S 22 = M 23 ( M 23 + 1 ) M 22 ( M 22 + 1 ) 2 s_{23} = S_{23} - S_{22} = \dfrac {M_{23} (M_{23}+1) - M_{22} (M_{22}+1) } {2}

M 23 = 23 2 ( 23 + 1 ) = 276 M_{23} = \dfrac {23}{2} (23+1) = 276

M 22 = 22 2 ( 22 + 1 ) = 253 M_{22} = \dfrac {22}{2} (22+1) = 253

Therefore, s 23 = 276 × 277 253 × 254 2 = 76452 64262 2 = 12190 2 = 6095 s_{23} = \dfrac {276 \times 277 - 253 \times 254 } {2} = \dfrac {76452 - 64262} {2} = \dfrac {12190} {2} = \boxed{6095}

Very nice solution! Well done!

Stewart Feasby - 6 years, 8 months ago

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 n^{th} term formula?

Stewart Feasby - 6 years, 8 months ago

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1 2 n ( n 2 + 1 ) \dfrac { 1 }{ 2 } n\left( { n }^{ 2 }+1 \right)

Michael Mendrin - 6 years, 8 months ago

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Absolutely correct! Well done

Stewart Feasby - 6 years, 8 months ago
Prabhat Routray
Oct 12, 2014

the sum of nth term in this series is = (n^3+n)/2

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