જો sum_{r=1}^{n} T_{r} = frac{(2n-1)(2n+1)(2n | vedclass

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(C) આપેલ છે $S_{n} = \sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$.$T_{n} = S_{n} - S_{n-1} = \frac{(2n-1)(2n+1)(2n+3)(2n+5) - (2n-3)(2n

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(C) આપેલ છે $S_{n} = \sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$.$T_{n} = S_{n} - S_{n-1} = \frac{(2n-1)(2n+1)(2n+3)(2n+5) - (2n-3)(2n-1)(2n+1)(2n+3)}{64}$.$T_{n} = \frac{(2n-1)(2n+1)(2n+3) [ (2n+5) - (2n-3) ]}{64} = \frac{(2n-1)(2n+1)(2n+3) 	imes 8}{64} = \frac{(2n-1)(2n+1)(2n+3)}{8}$.તેથી,$\frac{1}{T_{r}} = \frac{8}{(2r-1)(2r+1)(2r+3)}$.આંશિક અપૂર્ણાંકનો ઉપયોગ કરતા: $\frac{1}{(2r-1)(2r+1)(2r+3)} = \frac{1}{4} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} ight)$.તેથી,$\sum_{r=1}^{n} \frac{1}{T_{r}} = 8 	imes \frac{1}{4} \sum_{r=1}^{n} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} ight) = 2 \left( \frac{1}{1 	imes 3} - \frac{1}{(2n+1)(2n+3)} ight)$.$n ightarrow \infty$ લેતા,બીજું પદ શૂન્ય થઈ જશે.પરિણામ $= 2 	imes \frac{1}{3} = \frac{2}{3}$.

જો $\sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$ હોય,તો $\lim_{n \rightarrow \infty} \sum_{r=1}^{n} \left(\frac{1}{T_{r}}\right)$ ની કિંમત શોધો:

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ધન પૂર્ણાંકો $n$ માટે,જો $4 a_n = (n^2 + 5n + 6)$ અને $S_n = \sum_{k=1}^n \left(\frac{1}{a_k}\right)$ હોય,તો $507 S_{2025}$ નું મૂલ્ય કેટલું થાય?

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શ્રેણીનો સરવાળો શોધો: $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{n(n + 1)}$

શ્રેણી $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \ldots$ ના $n$ પદોનો સરવાળો શોધો.

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