શ્રેણી frac{3}{1 cdot 2} cdot frac{1}{2} + fr | vedclass

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Question

(A) $n$-મું પદ $t_n = \frac{n+2}{n(n+1)} \cdot \left( \frac{1}{2} \right)^n$ છે. આપણે $\frac{n+2}{n(n+1)} = \frac{2(n+1) - n}{n(n+1)} = \frac{2}{n} -

Vedclass · Accepted Answer

$1 - \frac{1}{(n + 1) 2^n}$

Vedclass · Answer

(A) $n$-મું પદ $t_n = \frac{n+2}{n(n+1)} \cdot \left( \frac{1}{2} \right)^n$ છે. આપણે $\frac{n+2}{n(n+1)} = \frac{2(n+1) - n}{n(n+1)} = \frac{2}{n} - \frac{1}{n+1}$ લખી શકીએ. તેથી,$t_n = \left( \frac{2}{n} - \frac{1}{n+1} \right) \left( \frac{1}{2} \right)^n = \frac{1}{n} \left( \frac{1}{2} \right)^{n-1} - \frac{1}{n+1} \left( \frac{1}{2} \right)^n$. ધારો કે $f(n) = \frac{1}{n} \left( \frac{1}{2} \right)^{n-1}$. તો $t_n = f(n) - f(n+1)$. સરવાળો $S_n = \sum_{k=1}^n t_k = f(1) - f(n+1)$. $f(1) = 1$ અને $f(n+1) = \frac{1}{(n+1) 2^n}$. તેથી,$S_n = 1 - \frac{1}{(n+1) 2^n}$.

શ્રેણી $\frac{3}{1 \cdot 2} \cdot \frac{1}{2} + \frac{4}{2 \cdot 3} \cdot \left( \frac{1}{2} \right)^2 + \frac{5}{3 \cdot 4} \cdot \left( \frac{1}{2} \right)^3 + \dots$ ના $n$ પદ સુધીનો સરવાળો શોધો.

Similar Questions

$\sum_{k=0}^{12} \frac{1}{\sin \left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right) \sin \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)} = $

$\frac{{\frac{1}{2} \cdot \frac{2}{2}}}{{{1^3}}} + \frac{{\frac{2}{2} \cdot \frac{3}{2}}}{{{1^3} + {2^3}}} + \frac{{\frac{3}{2} \cdot \frac{4}{2}}}{{{1^3} + {2^3} + {3^3}}} + \dots + n \text{ પદો} =$

શ્રેણી $\frac{1}{{1 + {1^2} + {1^4}}} + \frac{2}{{1 + {2^2} + {2^4}}} + \frac{3}{{1 + {3^2} + {3^4}}} + \dots$ ના $n$ પદોનો સરવાળો શોધો.

જો $\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \dots (n \text{ પદો}) = \frac{k n}{n+1}$ હોય,તો $k$ ની કિંમત શોધો.

જો $S_n = \frac{n(n + 1)(n + 2)}{6}$ હોય,તો $\sum_{n = 1}^\infty \frac{1}{t_n} = $

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