Let = _ r=0 ^n (4 r^2+2 r+1 ) ^n C_r and = ( _ r=0 ^n ^n C_r r+1 )+ 1 n+1 . If 140 < 2 < 281 , then the value of n is _______

Step-by-step solution

aligned & = _ r=0 ^n (4 r^2+2 r+1 ) ^n C_r & =4 _ r=0 ^n r^2 n r n-1 r-1 +2 _ r=0 ^n r n r n-1 r- + _ r=0 ^n ^n C_r & +4 n _ r=0 ^n n-1 r-1 +2 n _ r=0 ^n n-1 r-1 + _ r=0 ^n ^n C_r & =4 n(n-1) 2^ n-2 +4 n 2^ n-1 +2 n 2^ n-1 +2^n & =2^ n-2 [4 n(n-1)+8 n+4 n+4] & =2^ n-2 [4 n^2+8 n+4 ] & =2 n(n+1)^2 & = _ r=0 ^n ^n C_r r+1 + 1 n+1 & = _ r=0 ^n C^ n+1 C_ r+1 n+1 + 1 n+1 & = 1 n+1 (1+ ^ n+1 C₁+ .+ n+1 n+1 ) & = 2^ n+1 n+1 & 2 = 2^ n+1 (n+1)^2 2^ n+1 (n+1)=(n+1)^3 & 140 < (n+1)^3 < 281 & n=4 (n+1)^3=125 & n=5 (n+1)^3=216