Find Sum!!

You may also note that by the Method of Differences that the difference are in G.P. , so you can write the $T_r$ of the sequence as $T_{r} = a\cdot r^{n-1} + b\cdot n + c$ .

Now , to determine the constants $a,b$ and $c$ we put $n=1,2,3 , \dots$ and equate them with the corresponding terms of the given series .

Which comes out to be : $a=2,b=0,c=-1$

$T_r = 2^{r} - 1 \\ \rightarrow S_n = \sum_{r=1}^{n} 2^{r} - \sum_{r=1}^{n} 1 \\= 2\cdot (2^{10} -1 ) - 10 = 2036$

$Let\\ S\quad =\quad 1\quad +\quad 3\quad +\quad 7\quad +\quad 15\quad +\quad 31+\quad ...\quad +\quad { T }_{ n }\quad --(1)\\ S\quad =\quad \quad \quad \quad \quad 1\quad +\quad 3\quad +\quad 7\quad +\quad 15\quad +\quad ...\quad +\quad { T }_{ n-1 }\quad +\quad { T }_{ n }\quad \quad --(2)\\ Perform\quad (1)\quad -\quad (2)\\ \therefore \quad 0\quad =\quad 1\quad +\quad (2\quad +\quad 4\quad +\quad 8\quad +\quad 16\quad +...\quad (n-1)\quad terms)\quad -\quad { T }_{ n }\\ \therefore \quad { T }_{ n }\quad =\quad 1\quad +\quad (2\quad +\quad 4\quad +\quad 8\quad +\quad 16\quad +...\quad (n-1)\quad terms)\\ \therefore \quad { T }_{ n }\quad =\quad 1\quad +\quad 2\quad *\quad \frac { ({ 2 }^{ n-1 }-1) }{ (2-1) } \quad =\quad { 2 }^{ n }-1\\ \therefore { S }_{ n }\quad =\quad \sum { { T }_{ n } } \quad =\quad \sum { ({ 2 }^{ n }-1) } \quad =\quad (2+{ 2 }^{ 2 }+{ 2 }^{ 3 }+...+{ 2 }^{ n })-n\\ \therefore \quad { S }_{ n }\quad =\quad { 2 }^{ (n+1) }-n-2\\ \therefore \quad { S }_{ 10 }\quad =\quad { 2 }^{ 11 }-10-2\quad =\quad 2036.\quad \\$

its a sequence who's difference are multiplied by 2