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- Question 1 Exercise 5.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... +2.2^2+3.2^3+4.2^4+\ldots$. ====Solution==== Let \begin{align} & S_n=1.2+2.2^2+3 \cdot 2^3+4 \cdot 2^4... end{align} Suburacting the (ii) from (i), we get \begin{align} (1-2) S_n&=1 \cdot 2+(2-1) 2^2+(3-2) 2^... 1+4 x+7 x^2+10 x^3+\ldots.$ ====Solution==== Let \begin{align} & S_n=1+4 x+7 x^2+10 x^3+\ldots +(3 n-2
- Question 1 Exercise 5.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... get $$A+B=0 \text{and} A=1$$ Putting $A=1$,then \begin{align}1+B&=0\\ B&=-1\end{align} hence \begin{align} & \dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1... ummation of the both sides of the above equation \begin{align} \sum_{k=1}^n T_k&=\sum_{k=1}^n(\dfrac{1
- Question 1 Exercise 5.1
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... of the both sides of the above equation, we get \begin{align}& \sum_{j=1}^n T_j=\sum_{j=1}^n(2 j-1)^2... T_2=1^2+2^2$, $T_3=1^2+2^2+3^2$ and so on we get \begin{align}& T_j=1^2+2^2+3^2+\ldots+j^2 \\ & =\dfra... ng of sum of the both sides of the above, we get \begin{align}\Rightarrow \sum_{j=1}^n T_j&=\dfrac{1}{
- Question 9 Review Exercise
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... ferences to compute the sum of the given series. \begin{align} & a_2-a_1=7-3=4 \\ & a_3-a_2=13-7=6 \\ ... 8, \ldots \end{align} Adding column wise, we get \begin{align} a_n-a_1&=4+6+8+\cdots+2 n \\ & =\dfrac{... w a_n&=n^2+n-2+a_1 \\ \Rightarrow a_n&=n^2+n-2+3 \because a_1=3 \\ \Rightarrow a_n&=n^2+n+1\end{align}
- Question 4 & 5 Exercise 5.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... ^2}+\dfrac{11}{3^3}+\ldots$ ====Solution==== Let \begin{align} & S_{\infty}=5+\dfrac{7}{3}+\dfrac{9}{3... end{align} Subtracting the (ii) from (i), we get \begin{align} & \dfrac{2}{3} S_{\infty}=5+\dfrac{2}{3... \dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}\end{align} $\because$ the series is geometric $r=\dfrac{1}{3}<1$
- Question 8 Review Exercise
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... a_n=n^3+3^n$$ Taking summation of the both sides \begin{align}\sum_{r=1}^n a_r&=\sum_{r=1}^n r^3+\sum_... n=2 n^2+3 n$$ Taking summation on the both sides \begin{align} \sum_{r=1}^n a_r&=2 \sum_{r=1}^n r^2+3 ... ====Solution==== The $n^{\text {th }}$ term is: \begin{align} & a_n=n(n+1)(n+4) \\ & a_n==n(n^2+5 n+4
- Question 2 & 3 Exercise 5.1
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... f the both sides from $j=1$ to $j=99$, we get $$ \begin{aligned} & \sum_{j=1}^{99} \tau_j=\sum_{j=1}^{... )+1)}{6}+\frac{99(99+1)}{2} \end{aligned} $$ $$ \begin{aligned} & =\frac{99(100)(199)}{6}+\frac{99(10... $1+3+5+\ldots+99$. So, first we find the total number of terms in the given series as: $$ \begin{align
- Question 7 & 8 Exercise 5.1
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... general term of the series is: $T_j=j(j+4)(j+8)$ \begin{align} & =j(j^2+12 j+32) \\ & =j^3+12 j^2+32 j... of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n j^3+12 ... king summation of the general term of the series \begin{align} & \sum_{j=1}^{2 n} T_j=4 \sum_{j=1}^{2
- Question 2 & 3 Exercise 5.2
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... 2 x^2+7^2 x^3+\ldots, x<1$. ====Solution==== Let \begin{align} & S_{\infty}=1+3^2 x+5^2 x^2+7^2 x^3+\l... \end{align} Subtracting the (2) from (2), we get \begin{align}& (1-x) S_{\infty}=1^2+(3^2-1^2) x+(5^2-... ots(4)\end{align} Again subtracting (4) from (3) \begin{align} & [(1-x)-x(1-x)] S_{\infty}=1+(8-1) x+(
- Question 1 Exercise 5.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... e series $4+13+28+49+76+\ldots$ ====Solution==== \begin{align} & a_2-a_1=13-4=9 \\ & a_3-a_2=28-13=15 ... \end{align} Which is in A.P. column wise, we get \begin{align} a_n-a_{n-1}&=9+15+21+\ldots+(n-1)\\ & =... a_n&=3(n^2-1)+a_1 \\ \Rightarrow a_n&=3 n^2-3+4 \because a_1=4 \\ \Rightarrow a_n&=3 n^2+1 \end{alig
- Question 2 Exercise 5.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... ries $4+14+30+52+80+114+\ldots$ ====Solution==== \begin{align} & a_2-a_1=14-4=10 \\ & a_3-a_2=30-14=16... lign} which is a A.P. Adding column wise, we get \begin{align} a_n-a_1&=10+16+22+\ldots+(n-1) \text { ... 2+n-4+a_1 \\ \Rightarrow a_n&=3 n^2+n-4+4 \quad \because a_1=4 \\ \Rightarrow a_n&=3 n^2+n\end{align}
- Question 3 Exercise 5.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... Solution==== We use the method of difference as: \begin{align} & a_2-a_1=10-4=6 \\ & a_3-a_2=18-10=8 \... dots$ which is a A.P. Adding column wise, we get \begin{align}& a_n-a_1=6+10+8-\ldots +(n-1) \text { t... 3 n-4+a_1 \\ & \Rightarrow a_n=n^2+3 n-4+4 \quad \because a_1=4 \\ & \Rightarrow a_n=n^2+3 n\end{align
- Question 5 Exercise 5.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... eries $3+9+21+45+93+189+\ldots$ ====Solution==== \begin{align} & a_2-a_1=9-3=6 \\ & a_3-a_2=21-9=12 \\... lign} which is a G.P. Adding column wise, we get \begin{align} a_n-a_1& =6+12+24+\ldots+(n-1) \text {t... _1 \\ \Rightarrow a_n&=6 \cdot 2^{n-1}-6+3 \quad \because a_1=3\\ \Rightarrow a_n&=3(2^n-1)\end{align}
- Question 6 Exercise 5.3
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... eries $28+32+52+152+652+\ldots$ ====Solution==== \begin{align} & a_2-a_1=32-28=4 \\ & a_3-a_2=52-32=20... h common ratio $r=5$. Adding column wise, we get \begin{align} a_n-a_1&=4+20+100+\ldots+(n-1) \text { ... 5^{n-1}-1 \\ \Rightarrow a_n&=5^{n-1}-1+28 \quad \because a_1=28 \\ \Rightarrow a_n&=5^{n-1}+27\end{al
- Question 2 & 3 Exercise 5.4
- ok of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Pesha... {k=1}^n \dfrac{1}{9 k^2+3 k-2}$ ====Solution==== \begin{align}\text { Let } S_n&=\sum_{k=1}^n \dfrac{1... -1}+\dfrac{B}{3 k+2}$$ Multiplying both sides by \begin{align} & (3 k-1)(3 k+2) \text { we get } \\ & ... {1}{3 k+2}]$$ Taking summation of the both sides \begin{align} \sum_{r=1}^n u_n&=\sum_{r=1}^n[\dfrac{1