Question 333 of 347MediumMCQ
Let n \[\ge\] 5 be an integer. If 9n - 8n - 1 = 64\[\alpha\] and 6n - 5n - 1 = 25\[\beta\], then \[\alpha\] - \[\beta\] is equal to
Solutions · 1
Official solution
Given,
\[{9^n} - 8n - 1 = 64\alpha \]
\[ \Rightarrow \alpha = {{{{(1 + 8)}^n} - 8n - 1} \over {64}}\]
\[ = {{\left( {{}^n{C_0}\,.\,1 + {}^n{C_1}\,.\,{8^1} + {}^n{C_2}\,.\,{8^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{8^n}} \right) - 8n - 1} \over {{8^2}}}\]
\[ = {{1 + 8n + {}^n{C_2}\,.\,{8^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{8^n} - 8n - 1} \over {{8^2}}}\]
\[ = {{{}^n{C_2}\,.\,{8^2} + {}^n{C_3}\,.\,{8^3}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{8^n}} \over {{8^2}}}\]
\[ = {}^n{C_2} + {}^n{C_3}\,.\,8 + {}^n{C_4}\,.\,{8^2}\,\, + \,\,.....\,\,{}^n{C_n}\,.\,{8^{n - 2}}\]
Also given,
\[{6^n} - 5n - 1 = 25\beta \]
\[ \Rightarrow \beta = {{{{(1 + 5)}^n} - 5n - 1} \over {25}}\]
\[ = {{{}^n{C_0}\,.\,1 + {}^n{C_1}\,.\,5 + {}^n{C_2}\,.\,{5^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{5^n} - 5n - 1} \over {{5^2}}}\]
\[ = {{1 + 5n + {}^n{C_2}\,.\,{5^2} + {}^n{C_3}\,.\,{5^3}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^2} - 5n - 1} \over {{5^2}}}\]
\[ = {{{}^n{C_2}\,.\,{5^2} + {}^n{C_3}\,.\,{5^3} + {}^n{C_4}\,.\,{5^4}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^n}} \over {{5^2}}}\]
\[ = {}^n{C_2} + {}^n{C_3}\,.\,5 + {}^n{C_4}\,.\,{5^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{5^{n - 2}}\]
\[\therefore\] \[\alpha - \beta \]
\[ = \left( {{}^n{C_2} + {}^n{C_3}\,.\,8 + {}^n{C_4}\,.\,{8^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{8^{n - 2}}} \right) - \left( {{}^n{C_2} + {}^n{C_3}\,.\,5 + {}^n{C_4}\,.\,{5^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^{n - 2}}} \right)\]
\[ = {}^n{C_3}\,.\,(8 - 5) + {}^n{C_4}\,.\,({8^2} - {5^2})\,\, + \,\,....\,\, + \,\,{}^n{C_n}({8^{n - 2}} - {5^{n - 2}})\]
Answer the question to unlock