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Question

If the sum to n terms of the series 312×22+522×32+732×42+ is 0.99, then the value of n is

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Solution

​​​​​​Let Tr be the rth term of the given series. Then,
Tr=(2r+1)r2(r+1)2,
=[1r21(r+1)2]
Hence, the required sum is
nr=1Tr=nr=1[1r21(r+1)2]
=(112122)+(122132)+( )+(1n21(n+1)2)
=11(n+1)2
=2n+n2(n+1)2
Now,
0.99=2n+n2(n+1)2
99100=2n+n2(n+1)299n2+99+198n=200n+100n2n2+2n99=0n=11,9
But nN
n=9

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