જો mathrm{b} એ mathrm{a} ની સાપેક | vedclass

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

Question

$(a-b)^{-1}+(a-2 b)^{-1}+\ldots \ldots+(a-n b)^{1}$

$=\frac{1}{a} \sum_{r=1}^{n}\left(1-\frac{r b}{a}\right)^{-1}$ $=\frac{1}{a} \sum_{r=1}^{n}\lef

Vedclass · Accepted Answer

$\frac{b^{2}}{3 a^{3}}$

Vedclass · Answer

$(a-b)^{-1}+(a-2 b)^{-1}+\ldots \ldots+(a-n b)^{1}$

$=\frac{1}{a} \sum_{r=1}^{n}\left(1-\frac{r b}{a}\right)^{-1}$ $=\frac{1}{a} \sum_{r=1}^{n}\left\{\left(1+\frac{r b}{a}+\frac{r^{2} b^{2}}{a^{2}}\right)+(\right.$ terms to be neglected $\left.)\right\}$

$=\frac{1}{a}\left[n+\frac{n(n+1)}{2} \cdot \frac{b}{a}+\frac{n(n+1)(2 n+1)}{6} \cdot \frac{b^{2}}{a^{2}}\right]$

$=\frac{1}{a}\left[n^{3}\left(\frac{b^{2}}{3 a^{2}}\right)+\ldots\right]$

So $\gamma=\frac{\mathrm{b}^{2}}{3 \mathrm{a}^{2}}$

Similar Questions

ધારો કે $\alpha=\sum_{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$ અને $\beta=\sum_{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$. છે. જો $5 \alpha=6 \beta$, હોય તો $n$=...........................

જો $\sum_{ r =0}^{10}\left(\frac{10^{ r +1}-1}{10^{ r }}\right) \cdot{ }^{11} C _{ r +1}=\frac{\alpha^{11}-11^{11}}{10^{10}}$ હોય તો $\alpha$ ની કિમંત મેળવો.

જો ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ... + {C_n}{x^n}$, તો ${C_0} + {C_2} + {C_4} + {C_6} + .....$ = . . .

${(1 + x + {x^2} + {x^3})^5}$ ના વિસ્તરણમાં $x$ ની યુગ્મ ઘાતકના સહગુણકનો સરવાળો મેળવો.