Given series is, 1×2125C0+2×3125C1+3×4125C2 +…+26×27125C25 ∵∫0x(1+x)25dx=∫0x[25C0+25C1x +25C2x2+…+25C25x25]dx
On integrating w.r.t. x, taking limits 0 to x, we get [26(1+x)26]0x =[25C0x+25C1⋅2x2+25C23x3+…+25C25⋅26x26]0x ⇒{261(1+x)26−261} =25C0x+25C1⋅2x2+…+25C25⋅26x26
Again, integrating w.r.t. x, taking limits 0 to 1 , we get 261∫01[1+x)26−1]dx =∫01[25C0x+25C1⋅2x2+…+25C2526x26]dx ⇒261[27(1+x)27−x]01 =[25C02x2+25C1⋅2×3x3+…+25C2526×27x27]01 ⇒261{27227−1−271}=2125C0+2×31⋅25C1 +…+26×271⋅25C25 ∴1×21⋅25C0+2×3125C1+3×4125C2 +…+26×271⋅25C25=26×27227−28