Question

найдите сумму чисел 5-15+45-...+405

Answer 1

$5-15+45-...+405 \\\ b_1=5 \\\ q= \frac{b_2}{b_1}= \frac{-15}{5}= -3 \\\ b_n=b_1q^{n-1} \\\ 5\cdot(-3)^{n-1}=405 \\\ (-3)^{n-1}=81 \\\ (-3)^{n-1}=(-3)^4 \\\ n-1=4 \\\ n=5 \\\ S_n= \frac{b_1(q^n-1)}{q-1} \\\ S_5= \frac{5\cdot ((-3)^5-1)}{-3-1} = \frac{5\cdot (-243-1)}{-4} = \frac{5\cdot 244}{4} = \frac{5\cdot 244}{4} =5\cdot 61=305$