Im sorry but is there any other way to do this as I have not learnt what ∑ stands for yet. Thanks
[MATH]\Sigma[/MATH] is the capital letter sigma in the Greek alphabet (the equivalent of S in the Roman alphabet) and means to
SUM (add up) (1+n-m) terms by substituting first m for k, then m+1 for k, and so on until you substitute n for k. So
[MATH]\sum_{k=1}^1 2^k = 2^1 = 2.[/MATH]
[MATH]\sum_{k=1}^2 2^k = 2^1 + 2^2 = 2 + 4 = 6.[/MATH]
[MATH]\sum_{k=0}^3 2^k = 2^0 + 2^1 + 2^2 + 2^3 = 1 + 2 + 4 + 8 = 15.[/MATH]
For a more complicated example that is pertinent to your question
[MATH]\left ( \sum_{k=0}^4 \dbinom{4}{k} * (-\ 1)^k * a^{(4-k)} * b^k \right) \equiv [/MATH]
[MATH]\left ( (-\ 1)^0 * \dbinom{4}{0} * a^{(4-0)} * b^0 \right ) + \left ( (-\ 1)^1 * \dbinom{4}{1} * a^{(4-1)} * b^1 \right ) + \left ( (-\ 1)^2 * \dbinom{4}{2} * a^{(4-2)} * b^2 \right ) + [/MATH]
[MATH]\left ( (-\ 1)^3 * \dbinom{4}{3} * a^{(4-3)} * b^3 \right ) + \left ( (-\ 1)^4 * \dbinom{4}{4} * a^{(4-4)} * b^4 \right ) =[/MATH]
[MATH]\dfrac{4! * a^4b^0}{0! * (4 - 0)!} -\dfrac{4! * a^3b^1}{1! * (4 - 1)!} + \dfrac{4! * a^2b^2}{2! * (4 - 2)!} - \dfrac{4! * a^1b^3}{3! * (4 - 3)!} + \dfrac{4! * a^1b^0}{4! * (4 - 0)!} = [/MATH]
[MATH]a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4 = (a - b)^4.[/MATH]