limx→∞121−x3+31+x2+521−x3+71+x2+....=

limx→∞ 121−x3+521−x3+……+31+x2+71+x2…..limx→∞ ∑k=1x (4k−3)21−x3+∑k=1x 4k−11+x2=16−3+42=−163+2=−103∑k=1xk2=x(x+1)(2x+1)6 ∑k=1xk=x(x+1)2