Search
You can find the results of your search below.
Fulltext results:
- Question 10 Exercise 7.3
- war, Pakistan. Q10 Find the sum of the following series: (i) $1-\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\ldots$ Solution: The given series is binomial series. Let it be identical with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 1+n ... } x^3+\ldots \end{aligned} $$ Comparing both the series, we have $n x=-\frac{1}{4}$ (I) and $\frac{n(n-1)
- Question 11 Exercise 7.3
- 3 !} \cdot \frac{1}{2^6}+\ldots $$ Now the above series is binomial series. Let it be identical with the exparsion of $(1+x)^{\prime \prime}$ that is $1+n x=\fra... {n(n-1(n-2))}{3 !} x^3+\ldots$ Comparing both the series, we have $n x=\frac{1}{2^2}=\frac{1}{4}$ (1) and
- Question 12 Exercise 7.3
- 5}{3 !} \cdot \frac{1}{2^6}+\ldots$ Now the above series is binomial series. Lel it be identical with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 10+n... } x^3+\ldots \end{aligned} $$ Comparing both the series, we have $n x=\frac{1}{2^2}=\frac{1}{4}.... (1)$
- Question 1 Exercise 7.1
- )$$ 3. For $n=k+1$, the $(k+1)^{t h}$ term of the series, which is: $$a_{k+1}=\mathbf{2}(k+1)=2 k+2 $$ Add
- Question 2 Exercise 7.1
- \end{align} 3. For $n=k+1$, the $k+1$ term of the series, which is: $$a_{k-1}=4(k+1)-3=4 k+1 $$ Adding thi
- Question 3 Exercise 7.1
- considering for $n=k+1$, the $(k+1)$ term of the series is $a_{k+1}=3(k+1)$. Adding this $(k+1)^{t h}$ t
- Question 4 Exercise 7.1
- considering for $n=k+1$, the $(k+1)$ term of the series is $a_{k+1}=4(k+1)-1$. Adding this $(k+1)^{t h}$
- Question 5 Exercise 7.1
- ned} 3. Now $n=k+1$ the $(k+1)$ term of the given series on left is $a_{k+1}=(k+1)^3$. Adding this $(k+1)
- Question 6 Exercise 7.1
- ign} 3. For $n=k+1$ the $(k+1)^{t h}$ term of the series on the left is $a_{k+1}=(k+1)[(k+1) !]$. Actding
- Question 7 Exercise 7.1
- ering for $n=k+1$, then $(k-1)^{t h}$ term of the series on left is $a_{k+1}=(k+1)(k+ 2)$. Adding this $(
- Question 8 Exercise 7.1
- ering for $n-k-1$, then $(k+1)^{t h}$ term of the series on the left is $a_{k+1}=2^k$. Adding this $a_{k+
- Question 9 Exercise 7.1
- ]$$ 3. For $n=k+1$, the $(k+1)^{t h}$ term of the series on left is $a_{k+1}=\frac{1}{3^{k+1}}$. Adding t
- Question 10 Exercise 7.1
- gn} 3. For $n=k+1$ then $(k+1)^{t h}$ term of the series on the left is $a_{k+1}=\left(\begin{array}{c}k+5
- Question 11 Exercise 7.1
- $ 3. For $n=k+1$ then $(k+i)^{u / i}$ term of the series on the left is $a_{k-1}=\left(\begin{array}{c}k-1