Question

The series of natural numbers is divided into groups 1 ; 2, 3, 4 ; 5, 6, 7, 8, 9, 10, 11 ; 12 to 26..... and so on. The sum of the numbers in the nth group is $a.2^{2n-1}-(2n+b)2^{n-1}+n+1$, then evaluate a + b.

Answer 1

Number of terms in each group 1, 3, 7, 15,..........
$lasttermineachgroup1,4,11,\mathrm{26............}{a}_n$
$NowS=1+4+11+26.............+a_n$
$S=1+4..............................a_n$
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$\Rightarrow a_n=1+3+7+15.......nterm$
$a_n=1+3+7+15...........nterm$
$a_n=1+3+7...............n^{th}term$
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$n^{th}term=1+2+4+8+............=(2^n-1)$
So $a_n={\sum }_{n=1}^n(2^n-1)=2(2^n-1)-n$
$Solasttermof{n}^{th}group=2(2^n-1)-n$
$lasttermof(n-1{)}^{th}group=2[2^{n-1}-1]-(n-1)$
$Sumofthenumbersinthe{n}^{th}group$
$=\{1+2+\mathrm{3......}\overline{2(2^n-1)-n}\}$
$-\{1+2+\mathrm{3......}\overline{2(2^{n-1}-1)-(n-1)}\}$
$\frac{(2^{n+1}-n-2)(2^{n+1}-n-1)-\{2^n-(n-1\left)\right\}(2^n-n)}{2}$
$={3.2}^{2n-1}-(2n+5)2^{n-1}+(n+1)$
$a+b=8$