∫(v1−vn)dv=∫(vn−1vn(1−vn))dvLetvn=tthennvn−1dv=dtorvn−1dv=(1n)dt∴I=(1n)∫(1t(1−t))dt=(1n)∫((11−t)+(1t))dt=(1n)(−log|1−t|+log|t|)+c=(1n)log∣∣(t1−y)∣∣+c=(1n)log∣∣(vn1−vn)∣∣+c
If α,β are the roots of x2−ax+b=0 and if αn+βn=Vn, then
Using Vn method, if Tn = (3n - 1)(3n + 2), Find the value of K if Vn - Vn−1 = K Tn __
[√(n-1)][√(n+1)]