Question 2 & 3 Exercise 5.1

Solutions of Question 2 & 3 of Exercise 5.1 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Q2 Find the sum $1.2+2.3+3.4+\ldots+99.100$. Solution: The given series is the product of the corresponding terms of the series $1+2+3+\ldots+99$ and $2+3+4+\ldots+100$, whose $n^{\text {th }}$ terms are $n(n+1)$ and the given series have 99 terms. Therefore, the $n^{\text {th }}$ term of the given series is: $\quad T_j=j(j+1)=j^2+j$. Taking sum of the both sides from $j=1$ to $j=99$, we get $$\begin{aligned} & \sum_{j=1}^{99} \tau_j=\sum_{j=1}^{99} j^2+\sum_{j=1}^{99} j \\ & =\frac{99(99+1)(2(99)+1)}{6}+\frac{99(99+1)}{2} \end{aligned}$$

$$\begin{aligned} & =\frac{99(100)(199)}{6}+\frac{99(100)}{2} \\ & =(33.50 .199)+(99.50) \\ & \Rightarrow 1.2+2.3+3.4+\ldots+99.100 \\ & =3368050 . \end{aligned}$$

Q3 Find the sum $1^2+3^2+5^2+\ldots+99^2$. Solution: The each term of the given series is the square of the term of the series $1+3+5+\ldots+99$. So, first we find the total number of terms in the given series as: $$\begin{aligned} & a_j=99=1+2(j-1) \\ & \Rightarrow 2 j-1=99 \\ & \Rightarrow j=\frac{100}{2}=50 . \end{aligned}$$ The sum of the 50 terms of the series $$\begin{aligned} & \sum_{j=1}^{j=50} T_j=\sum_{j=1}^{j=50}(2 j-1)^2 \\ & =4 \sum_{j=1}^{50} j^2-4 \sum_{j=1}^{50} j+\sum_{j=1}^{50} 1 \\ & =4 \frac{49(49+1)(2(49)+1))}{6} \\ & -4 \frac{49(49+1)}{2}+50 \\ & =\frac{4(49)(50)(99)}{6}-\frac{4(49)(50)}{2}+50 \\ & =161700-4900+50 \Rightarrow 1^2+3^2+5^2+ \\ & \ldots+99^2=156850 . \end{aligned}$$

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