# Question 1 Exercise 5.1

Solutions of Question 1 of Exercise 5.1 of Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Sum the series $1^2+3^2+5^2+7^2+\ldots$ up to $n$ terms.

We see that each term of the given series is square of the terms of the series $1+3+5+\ldots$

whose $n^{\text {th }}$ term is $2 n-1$.

Therefore $n^{t h}$-term of - the given series is: $$T_j=(2 j-1)^2$$.

Taking summation 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 \\ & =\sum_{j=1}^n(4 j^2-4 j+1)\\ & =4 \sum_{j=1}^n j^2-4 \sum_{j=1}^n j+\sum_{j=1}^n 1 \\ & =4 \dfrac{n(n+1)(2 n+1)}{6}-4 \dfrac{n(n+1)}{2}+n \\ & =n[\dfrac{4(n+1)(2 n+1)-12(n+1)+6}{6}] \\ & =n[\dfrac{4(2 n^2+3 n+1)-12 n-12+6}{6}] \\ & =n[\dfrac{8 n^2+12n+4-12 n-6}{6}] \\ & =n[\dfrac{8 n^2-2}{6}] \\ & =\dfrac{n(4 n^2-1)}{3}\end{align}

Sum the series $1^2+(1^2+2^2)+(1^2+2^2+3^2)+\ldots$ up to $n$ terms.

In the given series, we see that $T_1=1^2, T_2=1^2+2^2$, $T_3=1^2+2^2+3^2$ and so on we get \begin{align}& T_j=1^2+2^2+3^2+\ldots+j^2 \\ & =\dfrac{j(j+1)(2 j+1)}{6}\end{align} Taking of sum of the both sides of the above, we get \begin{align}\Rightarrow \sum_{j=1}^n T_j&=\dfrac{1}{6} \sum_{j+1}^{j=n} j(2 j^2+3 j+1) \\ & =\dfrac{1}{6} \sum_{j=1}^n(2 j^3+3 j^2+j) \\ & =\dfrac{1}{6}[2 \sum_{j=1}^n j^3 +3 \sum_{j=1}^n j^2+\sum_{j+1}^n j] \\ & =\dfrac{1}{6}[2(\dfrac{n(n+1)}{2})^2+3 \dfrac{n(n+1)(2 n+1)}{6}+\dfrac{n(n+1)}{2}] \\ & =\dfrac{n(n+1)}{12}[n(n+1)+\dfrac{3(2 n+1)}{3}+1] \\ & =\dfrac{n(n+1)}{12}[\dfrac{3 n(n+1)+3(2 n+1)+3}{3}] \\ & =\dfrac{n(n+1)}{36}[3 n^2+9 n+6] \\ & =\dfrac{n(n+1)}{12}[n^2+3 n+2] \\ & =\dfrac{n(n+1)}{12}[n^2+n+2 n+2]\\ & =\frac{n(n+1)}{12}[n(n+1)+2(n+1)] \\ & =\dfrac{n(n+1)^2(n+2)}{12}\end{align}

Sum the series $2^2+4^2+6^2+8^2+\ldots$ up to $n$ terms.

The $n^{t h}$-term of the series is: $T_j=(\dot{2} j)^2=4 j^2$

Taking sum of the both sides, we get \begin{align}& \sum_{j=1}^n T_j=4 \sum_{j=1}^{j=n} j^2 \\ & =4 \dfrac{n(n+1)(2 n+1)}{6} \\ & =\dfrac{2 n(n+1)(2 n+1)}{3}\end{align}

Sum the series $1^3+3^3+5^3+7^3+\ldots$ up to $n$ terms. (iv) $1^3+3^3+5^3+7^3+\ldots$

The $n^{t h}$-term of the series is: $$T_1=(2 j-1)^3=8 j^3-12 j^2+6 j-1$$ Taking sum of the both sides of the above, we get \begin{align}& \sum_{j=1}^{11} T_j=8 \sum_{j=1}^n j^3-12 \sum_{i=1}^n j^2 \\ & +6 \sum_{j=1}^n j-\sum_{j=1}^n 1 \\ & =8(\dfrac{n(n+1)}{2})^2-12 \dfrac{n(n+1)(2 n+1)}{6} \\ & +6 \dfrac{n(n+1)}{2}-n \\ & =8 \dfrac{n^2(n+1)^2}{4}-2 n(n+1)(2 n+1) +3 n(n+1)-n \\ & =2 n^2(n+1)^2-2 n(n+1)(2 n+1)+3 n(n+1)-n \\ & =n(n+1)[2 n(n+1)-2(2 n+1)+3]-n\\ & =n(n+1)[2 n^2+2 n-4n-2+3]-n \\ & =n(n+1)[2 n^2-2 n+1]-n \\ & =n[(n+1)(2 n^2-2 n+1)-1] \\ &= n(2n^3-2n^2+n+2n^2-2n+1-1)\\ & =n(2 n^3-n)\end{align}

Sum the series $1^3+5^3+9^3+\ldots$ up to $n$ terms.

Her each term of the series is cube of the each term of the series $1+5+9+\ldots$,

whose $n^{\text {th }}$ term is: $a_j=1+4(j-1)=4 j-3$. \begin{align}& \therefore T_j-(4 j-3)^3 \\ & =64 j^3-144 j^2+108 j-27\end{align} Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=64 \sum_{j=1}^n j^3-144 \sum_{j=1}^n j^2 \\ & +108 \sum_{j=1}^n j-27 \sum_{j=1}^n 1 \\ & =64(\dfrac{n(n+1)}{2})^2-144 \dfrac{n(n+1)(2 n+1)}{6}+108 \dfrac{n(n-1)}{2}-27 n \\ & =64 \dfrac{n^2(n+1)^2}{4}-24 n(n+1)(2 n+1)+54 n(n+1)-27 n \\ & =n [ 16 n(n^2+2 n+1)-24(2 n^2+3 n+1)+54(n+1)-27] \\ & =n[16 n^3+32 n^2+16 n-48 n^2-72 n-24+54 n+54-27] \\ & =n[16 n^3-16 n^2-2 n+3]\end{align}