AI-তৈরি নমুনা উত্তর
প্রদত্ত ধারা,S = 1 3 + 1 ⋅ 3 3 ⋅ 6 + 1 ⋅ 3 ⋅ 5 3 ⋅ 6 ⋅ 9 + 1 ⋅ 3 ⋅ 5 ⋅ 7 3 ⋅ 6 ⋅ 9 ⋅ 12 + ⋯ ∞ S = \frac{1}{3} + \frac{1\cdot 3}{3\cdot 6} + \frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9} + \frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12} + \cdots \infty S = 3 1 + 3 ⋅ 6 1 ⋅ 3 + 3 ⋅ 6 ⋅ 9 1 ⋅ 3 ⋅ 5 + 3 ⋅ 6 ⋅ 9 ⋅ 12 1 ⋅ 3 ⋅ 5 ⋅ 7 + ⋯ ∞
দ্বিপদী বিস্তৃতি ( 1 − x ) − n = 1 + n x + n ( n + 1 ) 2 ! x 2 + n ( n + 1 ) ( n + 2 ) 3 ! x 3 + ⋯ (1 - x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \cdots ( 1 − x ) − n = 1 + n x + 2 ! n ( n + 1 ) x 2 + 3 ! n ( n + 1 ) ( n + 2 ) x 3 + ⋯ 1 + S 1 + S 1 + S n x = 1 3 … ( 1 ) nx = \frac{1}{3} \quad \dots (1) n x = 3 1 … ( 1 ) n ( n + 1 ) 2 ! x 2 = 1 ⋅ 3 3 ⋅ 6 = 1 6 … ( 2 ) \frac{n(n+1)}{2!}x^2 = \frac{1\cdot 3}{3\cdot 6} = \frac{1}{6} \quad \dots (2) 2 ! n ( n + 1 ) x 2 = 3 ⋅ 6 1 ⋅ 3 = 6 1 … ( 2 )
সমীকরণ (2)-কে (1) দ্বারা ভাগ করে পাই:n ( n + 1 ) x 2 2 n x = 1 / 6 1 / 3 \frac{n(n+1)x^2}{2nx} = \frac{1/6}{1/3} 2 n x n ( n + 1 ) x 2 = 1/3 1/6 ⟹ ( n + 1 ) x 2 = 1 2 \implies \frac{(n+1)x}{2} = \frac{1}{2} ⟹ 2 ( n + 1 ) x = 2 1 ⟹ n x + x = 1 \implies nx + x = 1 ⟹ n x + x = 1 ⟹ 1 3 + x = 1 \implies \frac{1}{3} + x = 1 ⟹ 3 1 + x = 1 ⟹ x = 1 − 1 3 = 2 3 \implies x = 1 - \frac{1}{3} = \frac{2}{3} ⟹ x = 1 − 3 1 = 3 2
সমীকরণ (1) হতে পাই:n ( 2 3 ) = 1 3 ⟹ n = 1 2 n\left(\frac{2}{3}\right) = \frac{1}{3} \implies n = \frac{1}{2} n ( 3 2 ) = 3 1 ⟹ n = 2 1
সুতরাং,1 + S = ( 1 − x ) − n = ( 1 − 2 3 ) − 1 2 = ( 1 3 ) − 1 2 = 3 1 + S = (1 - x)^{-n} = \left(1 - \frac{2}{3}\right)^{-\frac{1}{2}} = \left(\frac{1}{3}\right)^{-\frac{1}{2}} = \sqrt{3} 1 + S = ( 1 − x ) − n = ( 1 − 3 2 ) − 2 1 = ( 3 1 ) − 2 1 = 3 ⟹ S = 3 − 1 \implies S = \sqrt{3} - 1 ⟹ S = 3 − 1
উত্তর: 3 − 1 \sqrt{3} - 1 3 − 1