Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?
Speedy Summations
Age 16 to 18 Challenge Level
In the video below, Alison works out \(\sum_{i=1}^{10} i\).
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Alison writes out $\sum_{i=1}^{10} i = 1+2+3+4+5+6+7+8+9+10.$
Next, Alison writes the numbers from 1 to 10, and then the same set of numbers in decreasing order, 10 to 1, underneath, then adds them in pairs. This gives $10 \times 11=110$.
Finally Alison writes the answer $55$ next to the original sum.
How could you adapt this method to work out the following sums?
$\sum_{i=1}^{100} i$
$2+4+6+\dots+96+98+100$
$\sum_{k=1}^{20} (4k+12)$
$37+42+47+52+\dots+102+107+112$
The sum of the first $n$ terms of the sequence $a, (a+d), (a + 2d), (a + 3d) \dots$
After how many terms would $17+21+25+\dots$ be greater than $1000$?
Can you find the sum of all the integers less than $1000$ which are not divisible by $2$ or $3$?
Can you find a set of consecutive positive integers whose sum is 32?