Question

Consider the arithmetic sequence 4,12,20.a, Prove that the sum of consecutive terms of this sequence (starting from the first term) is always a perfect square.

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Answer 1

Step-by-step explanation:

d= 8

so t1= 4

t2= 12

t3= 20

t4= 28

and so on

t1= 4= 2^2

t1+t2 = 4+12= 16= 4^2

t1+t2+t3= 4+12+20= 36= 6^2

so we can say this will also continue ahead

hence proved

Answer 2

$\large\text{\underline{Solution:-}}$

The sequence has a common difference of 8, and the first term 4.

The formula for series of an arithmetic progression is,

$\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=\dfrac{n\{2a+(n-1)d\}}{2}$

where $a$ is the first term, $d$ is the common difference, and $n$ is the term number.

$\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=\dfrac{n\{8+8(n-1)\}}{2}$

$\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=4n^{2}$

$\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=(2n)^{2}$

So, the sum of the sequence is always a perfect square.

$\large\text{\underline{Maths activity:-}}$

Also, we can make another perfect square sum.

Since,

$\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}-\sum^{n-1}_{k=1}a_{k}=a_{n}\ (n\geq2)\text{ and }\sum^{1}_{k=1}a_{k}=a_{1}$

Put a perfect square in terms of $n$. We get,

$\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}=(kn)^{2}$

Then,

$\hookrightarrow\displaystyle\sum^{n}_{k=1}a_{k}-\sum^{n-1}_{k=1}a_{k}=(kn)^{2}-\{k(n-1)\}^{2}$

$\hookrightarrow\displaystyle a_{n}=(kn)^{2}-\{k(n-1)\}^{2}\ (\text{for }n\geq2)$

$\hookrightarrow a_{n}=k^{2}(n^{2}-n^{2}+2n-1)\ (\text{for }n\geq2)$

$\hookrightarrow a_{n}=k^{2}(2n-1)\ (\text{for }n\geq2)$

$a_{1}$ follows the sequence.

$\hookrightarrow a_{1}=k^{2}$

Choosing $k=1$ we get,

$\hookrightarrow a_{n}=2n-1\implies 1+3+5+7+\cdots+(2n-1)=n^{2}$

Now, we found that adding the first n terms of odd numbers obtain $n^{2}$. This also proves that the sum of the given sequence 4, 12, 20, …, 8n-4 obtain (2n)², it is a case where $k=2$.