Prove proposition 34 in the Maasei Hoshev 1 2 n 2 3

Prove proposition 34 in the Maasei Hoshev:

[(1 + 2 + ... + n) + (2 + 3 + ... + n) + ... + n] + [1 + (1 + 2) + ... + (1 + 2 + ... + (n - 1))] = n( 1 + 2 + ... + n).

Solution

solution

[(1 + 2 + ... + n) + (2 + 3 + ... + n) + ... + n] + [1 + (1 + 2) + ... + (1 + 2 + ... + (n - 1))] = n( 1 + 2 + ... + n)

think of sum in two halves.[(1 + 2 + ... + n) + (2 + 3 + ... + n) + ... + n] + [1 + (1 + 2) + ... + (1 + 2 + ... + (n - 1)) where first half fills in the second half.so that each term exactly 1+2+3+......+n that is 2+3+4+......+n is filled by 1, and 3+4+.........+n is filled by 1+2+..... and n is filled by (1+2+.....+(n-1)).

in this way we get n copies of (1+2+3+......+n)

so

[(1 + 2 + ... + n) + (2 + 3 + ... + n) + ... + n] + [1 + (1 + 2) + ... + (1 + 2 + ... + (n - 1))] = n( 1 + 2 + ... + n).