$(a)\;n^2+n\qquad(b)\;n^2-n+2\qquad(c)\;2n^2-2n+2\qquad(d)\;n^2-2n+2$

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- Sum of $n$ terms of an $A.P.$ with first term $a$ and common difference, $d$ is $\large\frac{n}{2}$$(2a+(n-1)d)$

Answer : (c) $\;2n^2-2n+2$

1. The numbers in the $n^{th}$ group are in $A.P.$ with common difference 1.

2. There are $2n-1$ numbers in $n^{th}$ group.

Step 1:

To find the first term in $n^{th}$ group:

The first term in each group are the sequence $ 1,2,5,10..........t_n$

$\therefore$ The first term of $n^{th}$ group is $n^{th}$ term $(t_n)$ of this sequence.

Let $S=1+2+5+10+................t_n+0$.........(i)

$\:\:\:\:\:S=\:\:0+1+2+5+10+..............+t_n$......(ii)

Subtracting (i)-(ii) we get

$0=1+(1+3+5+...........(n-1)terms)-t_n$

$\Rightarrow\:t_n=1+(1+3+5+.......(n-1)terms)$

$1+3+5+..........(n-1)terms$ is sum of $n-1$ terms of $A.P.$ with $d=2$

$=\large\frac{n-1}{2}$$(2\times1+(n-2)\times 2)=(n-1)(n-1)=(n-1)^2$

$\Rightarrow\:t_n=1+\large\frac{n-1}{2}$$(2+(n-2)2)$

$\Rightarrow\:t_n=1+(n-1)^2=n^2-2n+2$

$i.e., $ The first term of $n^{th}$ group is $n^2-2n+2$

Step 2:

To find the last term of the $n^{th}$ group:

We know that in $n^{th}$ group there are $2n-1$ terms and the numbers are in $A.P.$ with $d=1$

and with first term $t_n$

$\therefore $ The last term in the $n{th}$ group is $(2n-1)^{th}$ term in the series of $n^{th}$ group,

$i.e.,\:\:tn,(t_n+1),(t_n+2)...........$

$(2n-1)^{th}$ term of an $A.P.$ is $a+(2n-2)d$

which is $t_n+(2n-2)\times 1=n^2-2n+2+2n-2=n^2$

$\therefore$ The sum of first and last term of $n^{th}$ group is $ n^2-2n+2+n^2=2n^2-2n+2$

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