( [ _ n=3 ³² ⁻¹ (1+ _ k=1 ^n 2 k ) ]= )

Step-by-step solution

( aligned & _ k=1 ^n 2 k=n(n+1) & ( _ n=3 ³² ⁻¹ (1+ _ k=1 ^n 2 k ) ) & = ( _ n=3 ³² ⁻¹(1+n(n+1)) ) & = ( _ n=3 ³² ⁻¹ ( (n+1)-n 1+(n+1) n ) ) & _ n=3 ³² ⁻¹ ( (n+1)-n 1+(n+1) n )= _ n=3 ³² [ ⁻¹(n+1)- ⁻¹ n ] & = ( ⁻¹ 4- ⁻¹ 3+ ( ⁻¹ 5- ⁻¹ 4 )+ . . + ( ⁻¹ 33- ⁻¹ 34 ) . & = ⁻¹ 33- ⁻¹ 3= ⁻¹ [ 33-3 1+99 ]= ⁻¹ ( 3 10 ) & ( _ n=3 ³² ⁻¹ (n+1)-n 1+(n+1) n )= ⁻¹ [ ⁻¹ ( 3 10 ) ] & = ⁻¹ [ ⁻¹ ( 10 3 ) ]= 10 3 aligned ) Hence, option (1) is correct.