Question

Calculate 10, …+ 4n+6 for n= as odd number less than 12

Answer 1

If the terms are T₁, T₂, T₃, ..., T_n where T_n=4n+6, then we have the series:

10, 14, 18, ..., 4n+2, 4n+6. This is an arithmetic progression with first term a=10 and common difference d=4.

If we take the general AP as:

a, a+d, a+2d, ..., a+(n-1)d then we get the sum:

na+d(1+2+3+...+(n-1)).

The sum in parentheses is just the sum of the natural numbers up to N = N(N+1)/2. But we have N=n-1 so the sum is n(n-1)/2 whether n is odd or even.

So the ultimate sum S_n is na+dn(n-1)/2. a=10 so we have:

10n+2n(n-1)=10n+2n²-2n=2n²+8n or 2n(n+4).

EXAMPLE: n=3, S₃=2×3(3+4)=6×7=42=10+14+18; n=5, S₅=90=10+14+18+22+26.

Note that the formula also works for even n.

For n=1, 3, 5, 7, 9, 11 the sums are: 10, 42, 90, 154, 234, 330.