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sandesht

_{[r = 1 ]}

^{[n ]} r(r+1)(r+2)(r+3)(r+4).

Pl write detailed solution.

kap.mehra

See this solution by Sprinkle.........brilliant dude......simply awesome solution......

http://www.goiit.com/posts/list/algebra-how-to-find-the-sum-of-series-38498.htm#189470

only diff is that u have to take here:

r(r+1)(r+2)(r+3)(r+4) = (1/6) ( r(r+1)(r+2)(r+3)(r+4)(r+5) - (r-1)r(r+1)(r+2)(r+3)(r+4) )

and then proceed like what sprinkle has done.
millions of thanks to Sprinkle!!!

sprinkle

although kap.mehra has given the solution,

sandesht!

_{[r = 1 ]}

^{[n ]} r(r+1)(r+2)(r+3)(r+4) = 1.2.3.4.5 + 2.3.4.5.6 +...+ n(n+1)(n+2)(n+3)(n+4)

Tr = r(r+1)(r+2)(r+3)(r+4)

=> Tr = (1/6) [r(r+1)(r+2)(r+3)(r+4)(r+5) - (r-1)r(r+1)(r+2)(r+3)(r+4)]

=> T1 = (1/6) [1.2.3.4.5.6 - 0.1.2.3.4.5]

and T2 = (1/6) [2.3.4.5.6.7 - 1.2.3.4.5.6]

and T3 = (1/6) [3.4.5.6.7.8 - 2.3.4.5.6.7]
...........................................................
...........................................................
and Tn-1 = (1/6) [(n-1)n(n+1)(n+2)(n+3)(n+4) - (n-2)(n-1)n(n+1)(n+2)(n+3)]

and Tn = (1/6) [n(n+1)(n+2)(n+3)(n+4)(n+5) - (n-1)n(n+1)(n+2)(n+3)(n+4)]

adding all of them (first part of every term is canceled by second part on the following term)

Sn = (1/6) [n(n+1)(n+2)(n+3)(n+4)(n+5) - 0.1.2.3.4.5]

=> Sn = n(n+1)(n+2)(n+3)(n+4)(n+5)/6 ANS

kap.mehra

well the God himself is here :)

konichiwa2x

neat method sprinkle.... i wasted a lot of time doing it the usual way..this method is called the Vn method isnt it?

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