It is given that, expansion of (1+2x)n for x=1 is (1+2)n=6561 3n=6561 n=8 ‌ For ‌x=‌‌
1
√2
R=(1+2x)n =I+F (√2−1)8=F′ (√2+1)8+(√2−1)8=I+(F+F′) This implies, 2[(√2)8]+8C2(√2)6+8C4(√2)4+8C6(√2)2+8C8 =I+(F+F′) So F+F′=1 F=1−F′ =1−(√2−1)8 So 1−‌‌