1.Let n≥5n \geq 5n≥5 be an integer. If 9n−8n−1=64α9^{n} - 8n - 1 = 64\alpha9n−8n−1=64α and 6n−5n−1=25β6^{n} - 5n - 1 = 25\beta6n−5n−1=25β, then α−β\alpha - \betaα−β is equal toa.1+(n2)(8−5)+(n3)(82−52)+⋯+(nn)(8n−1−5n−1)1 + \binom{n}{2}(8-5) + \binom{n}{3}(8^{2} - 5^{2}) + \cdots + \binom{n}{n}(8^{n-1} - 5^{n-1})1+(2n)(8−5)+(3n)(82−52)+⋯+(nn)(8n−1−5n−1)b.1+(n2)(8−5)+(n4)(82−52)+⋯+(nn)(8n−2−5n−2)1 + \binom{n}{2}(8-5) + \binom{n}{4}(8^{2} - 5^{2}) + \cdots + \binom{n}{n}(8^{n-2} - 5^{n-2})1+(2n)(8−5)+(4n)(82−52)+⋯+(nn)(8n−2−5n−2)c.(n3)(8−5)+(n4)(82−52)+⋯+(nn)(8n−2−5n−2)\binom{n}{3}(8-5) + \binom{n}{4}(8^{2} - 5^{2}) + \cdots + \binom{n}{n}(8^{n-2} - 5^{n-2})(3n)(8−5)+(4n)(82−52)+⋯+(nn)(8n−2−5n−2)d.(n2)(8−5)+(n3)(82−52)+⋯+(nn)(8n−3−5n−3)\binom{n}{2}(8-5) + \binom{n}{3}(8^{2} - 5^{2}) + \cdots + \binom{n}{n}(8^{n-3} - 5^{n-3})(2n)(8−5)+(3n)(82−52)+⋯+(nn)(8n−3−5n−3)Login to continueOnly logged in users canattempt or see the solution.
