Bank Soal Matematika SMA Notasi Sigma

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Nilai dari i=441(4i3)\sum_{i=4}^{41}\left(4i-3\right) adalah....

A

3.360

B

3.306

C

3.603

D

3.036

E

6.306

Pembahasan:

menggunakan sifat operasi sumasi

  1. i=1nc=cn\sum_{i=1}^nc=cn
  2. i=1n(ai±bi)=i=1nai±i=1nbi\sum_{i=1}^n\left(a_i\pm b_i\right)=\sum_{i=1}^na_i\pm\sum_{i=1}^nb_i
  3. i=1ncai=ci=1nai \sum_{i=1}^nca_i=c\sum_{i=1}^na_{i\ }
  4. jika mm bilangan bulat dengan 1<m<n1<m<n dan aia_i menyatakan rumus suatu barisan bilangan, maka berlaku i=m+1nai=i=1nai i=1mai\sum_{i=m+1}^na_i=\sum_{i=1}^na_{i\ }-\sum_{i=1}^ma_i
  5. i=1ni=1+2+3+...+n=n(n+1)2\sum_{i=1}^ni=1+2+3+...+n=\frac{n\left(n+1\right)}{2}

maka

i=441(4i3)=i=4414ii=4413\sum_{i=4}^{41}\left(4i-3\right)=\sum_{i=4}^{41}4i-\sum_{i=4}^{41}3 (sifat 2)

                        =4i=441ii=4413\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\sum_{i=4}^{41}i-\sum_{i=4}^{41}3 (sifat 3)

                        =4(i=141ii=13i)(i=1413i=133)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\sum_{i=1}^{41}i-\sum_{i=1}^3i\right)-\left(\sum_{i=1}^{41}3-\sum_{i=1}^33\right) (sifat 4)

                        =4(41(41+1)23(3+1)2)(i=1413i=133)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\frac{41\left(41+1\right)}{2}-\frac{3\left(3+1\right)}{2}\right)-\left(\sum_{i=1}^{41}3-\sum_{i=1}^33\right) (sifat 5)

                        =4(41(41+1)23(3+1)2)(3(41)3(3))\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\frac{41\left(41+1\right)}{2}-\frac{3\left(3+1\right)}{2}\right)-\left(3\left(41\right)-3\left(3\right)\right) (sifat 1)

                        =4(41(42)23(4)2)(1239)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\frac{41\left(42\right)}{2}-\frac{3\left(4\right)}{2}\right)-\left(123-9\right)

                        =2(172212)(114)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\left(1722-12\right)-\left(114\right)

                        =2(1710)(114)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\left(1710\right)-\left(114\right)

                        =3420114\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3420-114

                        =3306\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3306

Jadi, nilai dari i=441(4i3)=3.306\sum_{i=4}^{41}\left(4i-3\right)=3.306

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