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If , then equals
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a
12n(n+1)
b
12n(n+2)
c
12n(n+3)
d
12n(n+5)
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Correct option is C
We have tn=Sn−Sn−1∀n≥2∴ tn=16n2n2+9n+13−16(n−1)2(n−1)2 =162n3−(n−1)3+9n2−(n−1)2 =166n2−6n+2+9(2n−1)+13 =166n2+12n+6 =(n+1)2Also, t1=S1=4=(1+1)2∴ ∑r=1n tr=∑r=1n (r+1)=12(n+1)(n+2)−1=12n(n+3)Similar Questions
Let be a positive integer, then is equal to :
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