1.

The value of 10∑n=1−2n∫−2n−1sin27x dx+10∑n=12n+1∫2nsin27x dx is equal to

Answer»

The value of 10∑n=1−2n∫−2n−1sin27x dx+10∑n=12n+1∫2nsin27x dx is equal to



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