frac{{frac{1}{2} cdot frac{2}{2}}}{{{1^3}}} + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $n$-મું પદ $T_n$ આ મુજબ છે: $T_n = \frac{{\frac{n}{2} \cdot \frac{n+1}{2}}}{{\sum_{k=1}^{n} k^3}} = \frac{{\frac{n(n+1)}{4}}}{{{{\left( {\frac{{n(

Vedclass · Accepted Answer

$\frac{n}{{n + 1}}$

Vedclass · Answer

(C) $n$-મું પદ $T_n$ આ મુજબ છે: $T_n = \frac{{\frac{n}{2} \cdot \frac{n+1}{2}}}{{\sum_{k=1}^{n} k^3}} = \frac{{\frac{n(n+1)}{4}}}{{{{\left( {\frac{{n(n + 1)}}{2}} \right)}^2}}}$ $= \frac{{\frac{{n(n + 1)}}{4}}}{{\frac{{{n^2}{{(n + 1)}^2}}}{4}}} = \frac{1}{{n(n + 1)}}$ આંશિક અપૂર્ણાંકનો ઉપયોગ કરતા: $T_n = \frac{1}{n} - \frac{1}{{n + 1}}$ સરવાળો $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{{k + 1}} \right)$ $S_n = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \dots + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)$ $S_n = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}}$.

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

Similar Questions

જો $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$ અને $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n$ હોય,તો બિંદુ $(m, n)$ કઈ રેખા પર આવેલું છે?

શ્રેણી $\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots$ ના પ્રથમ $n$ પદોનો સરવાળો શોધો.

જો $\frac{1}{2 \times 7} + \frac{1}{7 \times 12} + \frac{1}{12 \times 17} + \frac{1}{17 \times 22} + \dots$ $10$ પદો સુધી $= k$ હોય,તો $k =$

$\sum\limits_{r = 0}^{100} {(r^2 + 4r + 4)(r + 1)!}$ ની કિંમત :-

$\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{3}{{{r^2} - r + 9}}} \right)} $ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module