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.

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?