A) 1353300
B) 1253300
C) 1453300
D) 1753300
Correct Answer: A
Solution :
(a): \[1.3+3.5+5.7+......\]\[{{T}_{n}}={{T}_{n1}}\times {{T}_{n2}}\]\[=[1+(n-1)2]\times [3+(n-1)2]=(2n-1)(2n+1)\] \[\therefore {{T}_{n}}=4{{n}^{2}}-1\]. Also, \[\sum{{{T}_{n}}=\sum{(4{{n}^{2}}-1)=\sum{{{n}^{2}}-\sum{1}}}}\]\[=4\times \frac{n(n+1)(2n+1)}{6}-n\] \[=\frac{2n(n+1)(2n+1)-3n}{3}=\frac{n}{3}[2(n+1)(2n+1)-3]\] \[{{S}_{100}}=\sum{{{T}_{100}}=\frac{100}{3}[2\times 101\times 201-3]}\] \[=100\,[202\times 67-1]=100\,[13534-1]\] \[=100\times 13533=1353300\]
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