Hi Dhwani
5) See the pattern:
1×2×3 = (1/4) × [1×2×3×4  0×1×2×3]
2×3×4 = (1/4) × [2×3×4×5 1×2×3×4]
:::::::::::::::::::::::::::::::::::::::::::
n(n + 1)(n + 2) = (1/4) × [n(n + 1)(n + 2)(n + 3)  (n  1)n(n + 1)(n + 2)]
=> The required sum is = (1/4) × [n(n + 1)(n + 2)(n + 3)] = [(n² + 3n + 1)²  1]/4.
3) If a, b, c are in GP i.e. b² = ac
Taking log_{x} both sides we get,
2log_{x}b = log_{x}a + log_{x}c
=> 2/log_{b}x = 1/log_{a}x + 1/log_{c}x.
i.e. log_{a}x, log_{b}x and log_{c}x are in HP.
Kamal Lohia
