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Sum to 30 terms of the series is
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a
116465
b
4951984
c
435791
d
485791
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Correct option is B
tr=1r(r+1)(r+2)=12r−1r+1+12(r+2)[split into partial fractions] =121r−1r+1−121r+1−1r+2⇒ Sn=∑r=1n tr=121−1n+1−1212−1n+2=12nn+1−n4(n+2) Thus, S30=123031−304(32)=30464−31(31)(32)=4951984Similar Questions
Let be a positive integer, then is equal to :
799 666 8865
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