how to find the coefficient of x^100 in (1+x^2+x^4...)(1+x^3+x^6..)(1+x^5+x^10...) using binomial Th
You will have to calculate the coefficient of x100 in (1 + x2 + x4 + x6 + ...)(1 + x3 + x6 + ...)(1 + x5 + x10 + ...) the hard way, although it's not so hard as it seems. Notice the exponents of the last multiplicand here- 1 + x5 + x10 + x15 + ... x100. The coefficient of x are 0, 5, 10, ... 100. Therefore, we need to find the coefficient of x100, x95, x90.... in (1 + x2 + x4 + x6 + ...)(1 + x3 + x6 + ...). In other words, we need to find the number of solutions to 2x + 3y = 100, 2x + 3y = 95, 2x + 3y = 90, and so on. This can be found straight away through the formula given for the number of whole number solution of ax + by = c in the same article.
You will have to calculate the coefficient of x100 in (1 + x2 + x4 + x6 + ...)(1 + x3 + x6 + ...)(1 + x5 + x10 + ...) the hard way, although it's not so hard as it seems. Notice the exponents of the last multiplicand here- 1 + x5 + x10 + x15 + ... x100. The coefficient of x are 0, 5, 10, ... 100. Therefore, we need to find the coefficient of x100, x95, x90.... in (1 + x2 + x4 + x6 + ...)(1 + x3 + x6 + ...). In other words, we need to find the number of solutions to 2x + 3y = 100, 2x + 3y = 95, 2x + 3y = 90, and so on. This can be found straight away through the formula given for the number of whole number solution of ax + by = c in the same article.
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